-1/2 cos 2 x. By the method of inspection, look for a function whose derivative is sin 2x. The cosine is the natural candidate, and the chain rule supplies the inner factor: d/dx( cos 2x) = -2 sin 2x. Divide through by -2: d/dx(-1/2 cos 2x) = sin 2x. The factor 1/2 is exactly what compensates for the 2 the chain rule produces — that is the step to get right. ∫ sin 2x dx = -1/2 cos 2x + C.

1/3 sin 3 x. By inspection, differentiate the natural candidate sin 3x using the chain rule: d/dx( sin 3x) = 3 cos 3x. Divide by 3: d/dx(1/3 sin 3x) = cos 3x. So an anti derivative of cos 3x is 1/3 sin 3x, and ∫ cos 3x dx = 1/3 sin 3x + C.

1/2 e^2 x. By inspection, the exponential reproduces itself on differentiation, with the chain rule contributing the exponent's coefficient: d/dx(e^2x) = 2e^2x. Hence d/dx(1/2e^2x) = e^2x, so ∫ e^2x dx = 1/2e^2x + C.

1/(3 a)(a x+b)^3. Take a ≠ 0. By inspection, try the next power, (ax+b)^3, and differentiate by the chain rule: d/dx(ax+b)^3 = 3(ax+b)^2· a = 3a (ax+b)^2. Divide by 3a: d/dx(((ax+b)^3)/3a) = (ax+b)^2. The 1/a is the factor most often dropped here: it is the derivative of the inner function ax+b being undone. ∫ (ax+b)^2 dx = ((ax+b)^3)/3a + C (a ≠ 0). If a = 0 the function is the constant b^2, whose anti derivative is b^2x + C.

-1/2 cos 2 x-4/3 e^3 x. The anti derivative is linear, so treat the two terms separately. For the first term, as in Question 1, d/dx(-1/2 cos 2x) = sin 2x. For the second, d/dx(e^3x) = 3e^3x, so d/dx(1/3e^3x) = e^3x and therefore d/dx(-4/3e^3x) = -4e^3x. Adding, d/dx(-1/2 cos 2x - 4/3e^3x) = sin 2x - 4e^3x. Hence ∫ ( sin 2x - 4e^3x) dx = -1/2 cos 2x - 4/3e^3x + C.

4/3 e^3 x+x+ C. Use the linearity of the integral, ∫ (f + g) dx = ∫ f dx + ∫ g dx, together with ∫ e^kxdx = 1/ke^kx ( k ≠ 0 ) and ∫ 1 dx = x: ∫ (4e^3x + 1) dx = 4∫ e^3x dx + ∫ 1 dx = 4· (e^3x)/3 + x + C. So ∫ (4e^3x + 1) dx = 4/3e^3x + x + C.

∫ x^2(1-1/(x^2)) d x

(x^3)/3-x+ C. There is no product rule for integration, so simplify the integrand first (valid for x ≠ 0 ): x^2(1 - 1/(x^2)) = x^2 - 1. Now apply the power rule ∫ x^ndx = (x^n+1)/(n+1) (n ≠ -1): ∫ (x^2 - 1) dx = (x^3)/3 - x + C. Hence ∫ x^2(1 - 1/(x^2)) dx = (x^3)/3 - x + C, x ≠ 0.

(a x^3)/3+(b x^2)/2+c x+ C. By linearity, integrate term by term with the power rule ∫ x^ndx = (x^n+1)/(n+1) (here a, b, c are constants and come out of the integrals): ∫ (ax^2 + bx + c) dx = a∫ x^2dx + b∫ x dx + c∫ 1 dx. So ∫ (ax^2 + bx + c) dx = (a x^3)/3 + (b x^2)/2 + cx + C.

2/3 x^3+e^x+ C. By linearity, using the power rule for the first term and ∫ e^xdx = e^x for the second: ∫ (2x^2 + e^x) dx = 2∫ x^2dx + ∫ e^xdx = 2·(x^3)/3 + e^x + C. Hence ∫ (2x^2 + e^x) dx = 2/3x^3 + e^x + C.

∫(√ x -1/(√ x ))^2 d x

(x^2)/2+ log x -2 x+ C. Expand the square before integrating — the middle term is where the marks are lost: (√ x - 1/(√ x ))^2 = (√ x )^2 - 2·√ x ·1/(√ x ) + (1/(√ x ))^2 = x - 2 + 1/x. The presence of √ x and 1/(√ x ) restricts the domain to x 0. Now integrate term by term, using ∫ 1/xdx = log x: ∫ (x - 2 + 1/x) dx = (x^2)/2 - 2x + log x + C. So ∫ (√ x - 1/(√ x ))^2 dx = (x^2)/2 - 2x + log x + C, x 0.

∫ (x^3+5 x^2-4)/(x^2) d x

(x^2)/2+5 x+4/x+C. Divide each term of the numerator by x^2 first ( x ≠ 0 ): (x^3 + 5x^2 - 4)/(x^2) = x + 5 - 4/(x^2) = x + 5 - 4x^-2. Integrate term by term with the power rule; note ∫ x^-2dx = (x^-1)/-1 = -1/x, so the last term changes sign: ∫ (x + 5 - 4x^-2) dx = (x^2)/2 + 5x - 4(-1/x) + C. Hence ∫ (x^3 + 5x^2 - 4)/(x^2) dx = (x^2)/2 + 5x + 4/x + C, x ≠ 0.

∫ (x^3+3 x+4)/(√ x ) d x

2/7 x^7/2+2 x^3/2+8 √ x +C. Write √ x = x^1/2 and divide each term of the numerator by it, subtracting exponents ( x 0 ): (x^3 + 3x + 4)/(√ x ) = x^5/2 + 3x^1/2 + 4x^-1/2. Apply the power rule ∫ x^ndx = (x^n+1)/(n+1) to each term: ∫ x^5/2dx = (x^7/2)/(7/2) = 2/7x^7/2, 3∫ x^1/2dx = 3·(x^3/2)/(3/2) = 2x^3/2, 4∫ x^-1/2dx = 4·(x^1/2)/(1/2) = 8x^1/2. Adding, ∫ (x^3 + 3x + 4)/(√ x ) dx = 2/7x^7/2 + 2x^3/2 + 8√ x + C, x 0.

∫ (x^3-x^2+x-1)/x-1 d x

(x^3)/3+x+ C. Factorise the numerator by grouping so the denominator cancels: x^3 - x^2 + x - 1 = x^2(x - 1) + 1 (x - 1) = (x - 1)(x^2 + 1). Cancelling x - 1 is legitimate only where x ≠ 1, which is exactly where the integrand is defined: (x^3 - x^2 + x - 1)/(x - 1) = x^2 + 1, x ≠ 1. Integrate by the power rule: ∫ (x^2 + 1) dx = (x^3)/3 + x + C. So ∫ (x^3 - x^2 + x - 1)/(x - 1) dx = (x^3)/3 + x + C, x ≠ 1.

2/3 x^3/2-2/5 x^5/2+C. Multiply out first, since there is no product rule for integrals: (1 - x)√ x = x^1/2 - x^3/2. Apply the power rule to each term: ∫ x^1/2dx = (x^3/2)/(3/2) = 2/3x^3/2, ∫ x^3/2dx = (x^5/2)/(5/2) = 2/5x^5/2. Hence ∫ (1 - x)√ x dx = 2/3x^3/2 - 2/5x^5/2 + C, x ≥ 0.

∫ √ x (3 x^2+2 x+3) d x

6/7 x^7/2+4/5 x^5/2+2 x^3/2+ C. Distribute √ x = x^1/2 over the bracket, adding exponents: √ x (3x^2 + 2x + 3) = 3x^5/2 + 2x^3/2 + 3x^1/2. Integrate term by term with the power rule: 3·(x^7/2)/(7/2) = 6/7x^7/2, 2·(x^5/2)/(5/2) = 4/5x^5/2, 3·(x^3/2)/(3/2) = 2x^3/2. So ∫ √ x (3x^2 + 2x + 3) dx = 6/7x^7/2 + 4/5x^5/2 + 2x^3/2 + C, x ≥ 0.

∫(2 x-3 cos x+e^x) d x

x^2-3 sin x+e^x+ C. Integrate term by term, using the power rule, ∫ cos x dx = sin x and ∫ e^xdx = e^x: ∫ (2x - 3 cos x + e^x) dx = 2·(x^2)/2 - 3 sin x + e^x + C. Hence ∫ (2x - 3 cos x + e^x) dx = x^2 - 3 sin x + e^x + C.

∫(2 x^2-3 sin x+5 √ x ) d x

2/3 x^3+3 cos x+10/3 x^3/2+ C. Integrate term by term. The one to watch is the sine: ∫ sin x dx = - cos x, so the -3 sin x term integrates to +3 cos x. 2∫ x^2dx = 2/3x^3, -3∫ sin x dx = 3 cos x, 5∫ x^1/2dx = 5·(x^3/2)/(3/2) = 10/3x^3/2. Hence ∫ (2x^2 - 3 sin x + 5√ x ) dx = 2/3x^3 + 3 cos x + 10/3x^3/2 + C, x ≥ 0.

∫ sec x( sec x+ tan x) d x

tan x+ sec x+ C. Expand the product so that each piece is a standard derivative: sec x ( sec x + tan x) = sec^2x + sec x tan x. Now use the standard results d/dx( tan x) = sec^2x and d/dx( sec x) = sec x tan x, i.e. ∫ sec^2x dx = tan x, ∫ sec x tan x dx = sec x. Therefore ∫ sec x ( sec x + tan x) dx = tan x + sec x + C, valid on any interval on which cos x ≠ 0.

∫ ( sec^2 x)/( cosec^2 x) d x

tan x-x+ C. Convert to sines and cosines before integrating: ( sec^2x)/( cosec^2x) = (1/ cos^2x)/(1/ sin^2x) = ( sin^2x)/( cos^2x) = tan^2x. There is no direct formula for ∫ tan^2x dx, so use the identity 1 + tan^2x = sec^2x, i.e. tan^2x = sec^2x - 1: ∫ tan^2x dx = ∫ ( sec^2x - 1) dx = tan x - x + C. Hence ∫ ( sec^2x)/( cosec^2x) dx = tan x - x + C, valid on any interval on which cos x ≠ 0.

∫ (2-3 sin x)/( cos^2 x) d x

2 tan x-3 sec x+ C. Split the fraction over the common denominator cos^2x: (2 - 3 sin x)/( cos^2x) = 2/( cos^2x) - 3·( sin x)/( cos^2x) = 2 sec^2x - 3 sec x tan x, since ( sin x)/( cos^2x) = 1/( cos x)·( sin x)/( cos x) = sec x tan x. Using ∫ sec^2x dx = tan x and ∫ sec x tan x dx = sec x: ∫ (2 - 3 sin x)/( cos^2x) dx = 2 tan x - 3 sec x + C, valid on any interval on which cos x ≠ 0.

The anti derivative of (√ x +1/(√ x )) equals (A) 1/3 x^1/3+2 x^1/2+ C (B) 2/3 x^2/3+1/2 x^2+ C (C) 2/3 x^3/2+2 x^1/2+ C (D) 3/2 x^3/2+1/2 x^1/2+ C

C. An anti derivative of a function is any function whose derivative is that function, so compute the integral by the power rule ∫ x^ndx = (x^n+1)/(n+1) with n = 1/2 and n = -1/2: ∫ (√ x + 1/(√ x )) dx = ∫ (x^1/2 + x^-1/2) dx = (x^3/2)/(3/2) + (x^1/2)/(1/2) + C = 2/3x^3/2 + 2x^1/2 + C. Check by differentiating: d/dx(2/3x^3/2 + 2√ x ) = x^1/2 + x^-1/2, as required. This matches option (C). (C) 2/3x^3/2 + 2x^1/2 + C

If d/(d x) f(x)=4 x^3-3/(x^4) such that f(2)=0. Then f(x) is (A) x^4+1/(x^3)-129/8 (B) x^3+1/(x^4)+129/8 (C) x^4+1/(x^3)+129/8 (D) x^3+1/(x^4)-129/8

A. f is an anti derivative of the given derivative, so integrate, writing 3/(x^4) = 3x^-4: f(x) = ∫ (4x^3 - 3x^-4) dx = 4·(x^4)/4 - 3·(x^-3)/-3 + C = x^4 + 1/(x^3) + C. The sign is the trap: ∫ x^-4dx = (x^-3)/-3, and the minus in front of 3x^-4 cancels it, leaving +1/(x^3). Now use the condition f(2) = 0 to fix C: 2^4 + 1/(2^3) + C = 0 ⇒ 16 + 1/8 + C = 0 ⇒ C = -129/8. Therefore f(x) = x^4 + 1/(x^3) - 129/8, which is option (A).

log (1+x^2)+ C. Use the method of substitution: if t=g(x), then g'(x) dx=dt. Here the numerator is exactly the derivative of the denominator, so put t=1+x^2 ⇒ dt=2x dx. Then ∫2x/(1+x^2) dx= /t= log t + C. Since 1+x^2 0 for every real x, the modulus may be dropped. Resubstituting t=1+x^2, ∫2x/(1+x^2) dx= log (1+x^2)+ C.

1/3( log x )^3+ C. The factor 1/x is the derivative of log x, which invites the substitution t= log x ⇒ dt=1/x dx (x 0). Then ∫(( log x)^2)/x dx=∫ t^2 dt=(t^3)/3+ C. Resubstituting t= log x, ∫(( log x)^2)/x dx=(( log x)^3)/3+ C.

log 1+ log x +C. First take x common in the denominator — this is the step that makes the substitution visible: 1/(x+x log x)=1/(x (1+ log x)). Put t=1+ log x ⇒ dt=1/x dx (x 0). Then /(x(1+ log x))= /t= log t + C. Hence ∫1/(x+x log x) dx= log 1+ log x + C, x 0, log x≠-1.

cos ( cos x)+ C. The outer function has argument cos x, and sin x is (up to sign) its derivative. Put t= cos x ⇒ dt=- sin x dx, i.e. sin x dx=-dt. Watch the sign here — it is the usual slip. Then ∫ sin x sin ( cos x) dx=∫ sin t (-dt)=-∫ sin t dt= cos t+ C. Resubstituting t= cos x, ∫ sin x sin ( cos x) dx= cos ( cos x)+ C.

sin (a x+b) cos (a x+b)

-1/(4 a) cos 2(a x+b)+ C. Use the identity 2 sin θ cos θ= sin 2θ with θ=ax+b: sin (ax+b) cos (ax+b)=1/2 sin (2(ax+b) ). Therefore, using ∫ sin (kx+c) dx=-( cos (kx+c))/k with k=2a (a≠0), ∫ sin (ax+b) cos (ax+b) dx=1/2∫ sin (2ax+2b) dx=1/2(-( cos (2ax+2b))/2a)+ C. Hence ∫ sin (ax+b) cos (ax+b) dx=-( cos 2(ax+b))/4a+ C. (The substitution t= sin (ax+b) gives ( sin^2(ax+b))/2a+ C; the two answers differ only by the constant 1/4a.)

2/(3 a)(a x+b)^3/2+ C. Put t=ax+b ⇒ dt=a dx, i.e. dx=dt/a (a≠0). Then, by the power rule ∫ t^n dt=(t^n+1)/(n+1) with n= 12, ∫√ ax+b dx=1/a∫ t^1/2 dt=1/a·2/3t^3/2+ C. Hence ∫√ ax+b dx=2/3a(ax+b)^3/2+ C.

2/5(x+2)^5/2-4/3(x+2)^3/2+ C. The factor x outside is not the derivative of x+2, so substitute for the bracket and re-express x in the new variable — that back-substitution is the step to get right. Put t=x+2 ⇒ x=t-2, dx=dt. Then ∫ x√ x+2 dx=∫ (t-2) t^1/2 dt=∫(t^3/2-2t^1/2)dt=2/5t^5/2-4/3t^3/2+ C. Resubstituting t=x+2, ∫ x√ x+2 dx=2/5(x+2)^5/2-4/3(x+2)^3/2+ C.

1/6(1+2 x^2)^3/2+ C. Here x dx is proportional to the differential of 1+2x^2. Put t=1+2x^2 ⇒ dt=4x dx, i.e. x dx=dt/4. Then ∫ x√ 1+2x^2 dx=1/4∫ t^1/2 dt=1/4·2/3t^3/2+ C =1/6t^3/2+ C. Hence ∫ x√ 1+2x^2 dx=1/6(1+2x^2)^3/2+ C.

4/3(x^2+x+1)^3/2+ C. Notice that 4x+2=2(2x+1) and 2x+1 is the derivative of x^2+x+1. Put t=x^2+x+1 ⇒ dt=(2x+1) dx, so (4x+2) dx=2 dt. Then ∫(4x+2)√ x^2+x+1 dx=2∫ t^1/2 dt=2·2/3t^3/2+ C. Hence ∫(4x+2)√ x^2+x+1 dx=4/3(x^2+x+1)^3/2+ C.

2 log √ x -1 + C. A square root of x with x itself present: substitute for the root. Put t=√ x ⇒ x=t^2, dx=2t dt (x 0). Then x-√ x =t^2-t=t(t-1), and the factor t cancels: /(x-√ x )=∫(2t dt)/(t(t-1))=2 /t-1=2 log t-1 + C. Resubstituting t=√ x, ∫1/(x-√ x ) dx=2 log √ x -1 + C, x 0, x≠1.

2/3 √ x+4 (x-8)+ C. Put t=x+4 ⇒ x=t-4, dx=dt, so that (with x 0, hence t 4 0 ) /(√ x+4 ) dx= -4/(√ t ) dt=∫(t^1/2-4t^-1/2)dt=2/3t^3/2-8t^1/2+ C. Resubstituting t=x+4, /(√ x+4 ) dx=2/3(x+4)^3/2-8√ x+4 + C =2/3√ x+4 (x-8)+ C.

1/7(x^3-1)^7/3+1/4(x^3-1)^4/3+ C. Split the power of x so that one part supplies the differential: x^5 dx=x^3· x^2 dx. Put t=x^3-1 ⇒ dt=3x^2 dx, x^3=t+1. Then ∫(x^3-1)^1/3x^5 dx=1/3∫ t^1/3(t+1) dt=1/3∫(t^4/3+t^1/3)dt. Integrating, =1/3(3/7t^7/3+3/4t^4/3)+ C =1/7t^7/3+1/4t^4/3+ C. Hence ∫(x^3-1)^1/3x^5 dx=1/7(x^3-1)^7/3+1/4(x^3-1)^4/3+ C.

-1/(18(2+3 x^3)^2)+ C. Put t=2+3x^3 ⇒ dt=9x^2 dx, i.e. x^2 dx=dt/9. Then ∫(x^2)/((2+3x^3)^3) dx=1/9∫ t^-3 dt=1/9·(t^-2)/-2+ C =-1/(18t^2)+ C. Hence ∫(x^2)/((2+3x^3)^3) dx=-1/(18(2+3x^3)^2)+ C.

1/(x( log x)^m), x 0, m ≠ 1

(( log x)^1-m)/1-m+ C. Put t= log x ⇒ dt=1/x dx (x 0). Then /(x( log x)^m)=∫ t^-m dt. The power rule ∫ t^n dt=(t^n+1)/(n+1) applies because -m≠-1, which is exactly why the question stipulates m≠1: ∫ t^-m dt=(t^-m+1)/1-m+ C. Hence ∫1/(x( log x)^m) dx=(( log x)^1-m)/1-m+ C, x 0, m≠1.

-1/8 log 9-4 x^2 + C. The numerator is a constant multiple of the derivative of the denominator. Put t=9-4x^2 ⇒ dt=-8x dx, i.e. x dx=-dt/8. Then /(9-4x^2) dx=-1/8 /t=-1/8 log t + C. Since 9-4x^2 can be negative, the modulus must be kept: /(9-4x^2) dx=-1/8 log 9-4x^2 + C.

1/2 e^2 x+3+ C. Put t=2x+3 ⇒ dt=2 dx, i.e. dx=dt/2. Using ∫ e^t dt=e^t+ C, ∫ e^2x+3 dx=1/2∫ e^t dt=1/2e^t+ C. Hence ∫ e^2x+3 dx=1/2e^2x+3+ C.

- 1 2 e^ x^2 + C. Write the integrand as x e^ x^2 =x e^ -x^2 and put t=x^2 ⇒ dt=2x dx, i.e. x dx=dt/2. Then ∫ x e^ -x^2 dx=1/2∫ e^-t dt=1/2(-e^-t)+ C =-1/2e^-t+ C. Hence ∫ x e^ x^2 dx=- 1 2e^ x^2 + C.

e^ tan^-1 x + C. Recall d/dx( tan^-1x)=1/(1+x^2), so the factor dx/(1+x^2) is exactly the differential of the exponent. Put t= tan^-1x ⇒ dt=dx/(1+x^2). Then ∫ e^ tan^-1x 1+x^2 dx=∫ e^t dt=e^t+ C. Hence ∫ e^ tan^-1x 1+x^2 dx=e^ tan^-1x + C.

log (e^x+e^-x)+ C. Divide numerator and denominator by e^x — this is the step that turns the integrand into a derivative-over-function form: (e^2x-1)/(e^2x+1)=(e^x-e^-x)/(e^x+e^-x). Put t=e^x+e^-x ⇒ dt=(e^x-e^-x)dx. Then ∫(e^2x-1)/(e^2x+1) dx= /t= log t + C. Since e^x+e^-x 0 always, ∫(e^2x-1)/(e^2x+1) dx= log (e^x+e^-x)+ C.

(e^2 x-e^-2 x)/(e^2 x+e^-2 x)

1/2 log (e^2 x+e^-2 x)+ C. The numerator is, up to a factor 2, the derivative of the denominator: d/dx(e^2x+e^-2x)=2(e^2x-e^-2x). Put t=e^2x+e^-2x ⇒ dt=2(e^2x-e^-2x)dx. Then ∫(e^2x-e^-2x)/(e^2x+e^-2x) dx=1/2 /t=1/2 log t + C. Since e^2x+e^-2x 0, ∫(e^2x-e^-2x)/(e^2x+e^-2x) dx=1/2 log (e^2x+e^-2x)+ C.

1/2 tan (2 x-3)-x+C. There is no direct formula for ∫ tan^2θ dθ; convert it first using the identity sec^2θ=1+ tan^2θ, i.e. tan^2θ= sec^2θ-1, with θ=2x-3: ∫ tan^2(2x-3) dx=∫[ sec^2(2x-3)-1]dx. With t=2x-3, dt=2 dx, and ∫ sec^2t dt= tan t, =1/2 tan (2x-3)-x+ C. Hence ∫ tan^2(2x-3) dx=1/2 tan (2x-3)-x+ C.

-1/4 tan (7-4 x)+ C. Put t=7-4x ⇒ dt=-4 dx, i.e. dx=-dt/4. The negative sign from the coefficient of x is the usual slip here. Using ∫ sec^2t dt= tan t+ C, ∫ sec^2(7-4x) dx=-1/4∫ sec^2t dt=-1/4 tan t+ C. Hence ∫ sec^2(7-4x) dx=-1/4 tan (7-4x)+ C.

1/2( sin^-1 x)^2+ C. Recall d/dx( sin^-1x)= 1 √ 1-x^2, so put t= sin^-1x ⇒ dt= dx √ 1-x^2 (-1<x<1). Then ∫ sin^-1x √ 1-x^2 dx=∫ t dt=(t^2)/2+ C. Hence ∫ sin^-1x √ 1-x^2 dx=(( sin^-1x)^2)/2+ C, -1<x<1.

(2 cos x-3 sin x)/(6 cos x+4 sin x)

1/2 log 2 sin x+3 cos x +C. Differentiate the denominator and compare with the numerator: d/dx(6 cos x+4 sin x)=-6 sin x+4 cos x=2(2 cos x-3 sin x). So the numerator is exactly half the derivative of the denominator. Put t=6 cos x+4 sin x ⇒ dt=2(2 cos x-3 sin x) dx. Then ∫(2 cos x-3 sin x)/(6 cos x+4 sin x) dx=1/2 /t=1/2 log t + C. Hence ∫(2 cos x-3 sin x)/(6 cos x+4 sin x) dx=1/2 log 6 cos x+4 sin x + C, which (taking 2 out of the logarithm into the constant) may also be written as 1/2 log 3 cos x+2 sin x + C.

1/( cos^2 x(1- tan x)^2)

1/((1- tan x))+ C. Write 1/( cos^2x)= sec^2x, so that /( cos^2x(1- tan x)^2)=∫( sec^2x)/((1- tan x)^2) dx. Put t=1- tan x ⇒ dt=- sec^2x dx, i.e. sec^2x dx=-dt. Then ∫( sec^2x)/((1- tan x)^2) dx=- /(t^2)=-(-1/t)+ C =1/t+ C. Hence ∫1/( cos^2x(1- tan x)^2) dx=1/(1- tan x)+ C.

( cos √ x )/(√ x )

2 sin √ x + C. Put t=√ x ⇒ dt=1/(2√ x ) dx, i.e. dx/(√ x )=2 dt (x 0). Then ∫( cos √ x )/(√ x ) dx=2∫ cos t dt=2 sin t+ C. Hence ∫( cos √ x )/(√ x ) dx=2 sin √ x + C, x 0.

1/3( sin 2 x)^3/2+ C. Here cos 2x dx is proportional to the differential of sin 2x. Put t= sin 2x ⇒ dt=2 cos 2x dx, i.e. cos 2x dx=dt/2. Then ∫√ sin 2x cos 2x dx=1/2∫ t^1/2 dt=1/2·2/3t^3/2+ C =1/3t^3/2+ C. Hence ∫√ sin 2x cos 2x dx=1/3( sin 2x)^3/2+ C, valid on any interval where sin 2x≥0 (so that the square root is defined).

( cos x)/(√ 1+ sin x )

2 √ 1+ sin x + C. Put t=1+ sin x ⇒ dt= cos x dx. Then ∫( cos x)/(√ 1+ sin x ) dx= /(√ t )=∫ t^-1/2 dt=2t^1/2+ C. Hence ∫( cos x)/(√ 1+ sin x ) dx=2√ 1+ sin x + C, taken on an interval where 1+ sin x 0.

1/2( log sin x)^2+ C. Note that d/dx( log sin x)=( cos x)/( sin x)= cot x, so the whole factor cot x dx is the differential of the logarithm. Put t= log sin x ⇒ dt= cot x dx ( sin x 0). Then ∫ cot x log sin x dx=∫ t dt=(t^2)/2+ C. Hence ∫ cot x log sin x dx=(( log sin x)^2)/2+ C, sin x 0.

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NCERT Solutions · Class 12 Mathematics Integrals

261 exercises · 261 still being checked

EXERCISE 7.1 1–10 (part 1 of 27)

Find an anti derivative (or integral) of the following functions by the method of inspection.
Exercise 1
$\displaystyle \sin 2 x$
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NCERT’s answer
$\displaystyle -\frac{1}{2} \cos 2 x$
By the method of inspection, look for a function whose derivative is $\displaystyle \sin 2x$. The cosine is the natural candidate, and the chain rule supplies the inner factor: \[\frac{d}{dx}(\cos 2x) = -2\sin 2x.\] Divide through by $\displaystyle -2$: \[\frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x.\] The factor $\displaystyle \frac{1}{2}$ is exactly what compensates for the $\displaystyle 2$ the chain rule produces — that is the step to get right. \[\int \sin 2x\, dx = -\frac{1}{2}\cos 2x + C.\]
Exercise 2
$\displaystyle \cos 3 x$
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NCERT’s answer
$\displaystyle \frac{1}{3} \sin 3 x$
By inspection, differentiate the natural candidate $\displaystyle \sin 3x$ using the chain rule: \[\frac{d}{dx}(\sin 3x) = 3\cos 3x.\] Divide by $\displaystyle 3$: \[\frac{d}{dx}\left(\frac{1}{3}\sin 3x\right) = \cos 3x.\] So an anti derivative of $\displaystyle \cos 3x$ is $\displaystyle \frac{1}{3}\sin 3x$, and \[\int \cos 3x\, dx = \frac{1}{3}\sin 3x + C.\]
Exercise 3
$\displaystyle e^{2 x}$
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NCERT’s answer
$\displaystyle \frac{1}{2} e^{2 x}$
By inspection, the exponential reproduces itself on differentiation, with the chain rule contributing the exponent's coefficient: \[\frac{d}{dx}\left(e^{2x}\right) = 2e^{2x}.\] Hence \[\frac{d}{dx}\left(\frac{1}{2}e^{2x}\right) = e^{2x},\] so \[\int e^{2x}\, dx = \frac{1}{2}e^{2x} + C.\]
Exercise 4
$\displaystyle (a x+b)^{2}$
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NCERT’s answer
$\displaystyle \frac{1}{3 a}(a x+b)^{3}$
Take $\displaystyle a \neq 0$. By inspection, try the next power, $\displaystyle (ax+b)^{3}$, and differentiate by the chain rule: \[\frac{d}{dx}(ax+b)^{3} = 3(ax+b)^{2}\cdot a = 3a\,(ax+b)^{2}.\] Divide by $\displaystyle 3a$: \[\frac{d}{dx}\left(\frac{(ax+b)^{3}}{3a}\right) = (ax+b)^{2}.\] The $\displaystyle \frac{1}{a}$ is the factor most often dropped here: it is the derivative of the inner function $\displaystyle ax+b$ being undone. \[\int (ax+b)^{2}\, dx = \frac{(ax+b)^{3}}{3a} + C \qquad (a \neq 0).\] If $\displaystyle a = 0$ the function is the constant $\displaystyle b^{2}$, whose anti derivative is $\displaystyle b^{2}x + C$.
Exercise 5
$\displaystyle \sin 2 x-4 e^{3 x}$
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NCERT’s answer
$\displaystyle -\frac{1}{2} \cos 2 x-\frac{4}{3} e^{3 x}$
The anti derivative is linear, so treat the two terms separately. For the first term, as in Question $\displaystyle 1$, $\displaystyle \frac{d}{dx}\left(-\frac{1}{2}\cos 2x\right) = \sin 2x$. For the second, $\displaystyle \frac{d}{dx}\left(e^{3x}\right) = 3e^{3x}$, so $\displaystyle \frac{d}{dx}\left(\frac{1}{3}e^{3x}\right) = e^{3x}$ and therefore \[\frac{d}{dx}\left(-\frac{4}{3}e^{3x}\right) = -4e^{3x}.\] Adding, \[\frac{d}{dx}\left(-\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x}\right) = \sin 2x - 4e^{3x}.\] Hence \[\int \left(\sin 2x - 4e^{3x}\right) dx = -\frac{1}{2}\cos 2x - \frac{4}{3}e^{3x} + C.\]
Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:
Exercise 6
$\displaystyle \int\left(4 e^{3 x}+1\right) d x$
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NCERT’s answer
$\displaystyle \frac{4}{3} e^{3 x}+x+\mathrm{C}$
Use the linearity of the integral, $\displaystyle \int (f + g)\,dx = \int f\,dx + \int g\,dx$, together with $\displaystyle \int e^{kx}dx = \frac{1}{k}e^{kx}$ ($\displaystyle k \neq 0$) and $\displaystyle \int 1\,dx = x$: \[\int \left(4e^{3x} + 1\right) dx = 4\int e^{3x}\,dx + \int 1\,dx = 4\cdot \frac{e^{3x}}{3} + x + C.\] So \[\int \left(4e^{3x} + 1\right) dx = \frac{4}{3}e^{3x} + x + C.\]
Exercise 7
$\displaystyle \int x^{2}\left(1-\frac{1}{x^{2}}\right) d x$
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NCERT’s answer
$\displaystyle \frac{x^{3}}{3}-x+\mathrm{C}$
There is no product rule for integration, so simplify the integrand first (valid for $\displaystyle x \neq 0$): \[x^{2}\left(1 - \frac{1}{x^{2}}\right) = x^{2} - 1.\] Now apply the power rule $\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}\ (n \neq -1)$: \[\int \left(x^{2} - 1\right) dx = \frac{x^{3}}{3} - x + C.\] Hence \[\int x^{2}\left(1 - \frac{1}{x^{2}}\right) dx = \frac{x^{3}}{3} - x + C, \qquad x \neq 0.\]
Exercise 8
$\displaystyle \int\left(a x^{2}+b x+c\right) d x$
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NCERT’s answer
$\displaystyle \frac{a x^{3}}{3}+\frac{b x^{2}}{2}+c x+\mathrm{C}$
By linearity, integrate term by term with the power rule $\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}$ (here $\displaystyle a, b, c$ are constants and come out of the integrals): \[\int \left(ax^{2} + bx + c\right) dx = a\int x^{2}dx + b\int x\,dx + c\int 1\,dx.\] So \[\int \left(ax^{2} + bx + c\right) dx = \frac{a x^{3}}{3} + \frac{b x^{2}}{2} + cx + C.\]
Exercise 9
$\displaystyle \int\left(2 x^{2}+e^{x}\right) d x$
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NCERT’s answer
$\displaystyle \frac{2}{3} x^{3}+e^{x}+\mathrm{C}$
By linearity, using the power rule for the first term and $\displaystyle \int e^{x}dx = e^{x}$ for the second: \[\int \left(2x^{2} + e^{x}\right) dx = 2\int x^{2}dx + \int e^{x}dx = 2\cdot\frac{x^{3}}{3} + e^{x} + C.\] Hence \[\int \left(2x^{2} + e^{x}\right) dx = \frac{2}{3}x^{3} + e^{x} + C.\]
Exercise 10
$\displaystyle \int\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^{2} d x$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2}+\log |x|-2 x+\mathrm{C}$
Expand the square before integrating — the middle term is where the marks are lost: \[\left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} = (\sqrt{x})^{2} - 2\cdot\sqrt{x}\cdot\frac{1}{\sqrt{x}} + \left(\frac{1}{\sqrt{x}}\right)^{2} = x - 2 + \frac{1}{x}.\] The presence of $\displaystyle \sqrt{x}$ and $\displaystyle \frac{1}{\sqrt{x}}$ restricts the domain to $\displaystyle x > 0$. Now integrate term by term, using $\displaystyle \int \frac{1}{x}dx = \log|x|$: \[\int \left(x - 2 + \frac{1}{x}\right) dx = \frac{x^{2}}{2} - 2x + \log x + C.\] So \[\int \left(\sqrt{x} - \frac{1}{\sqrt{x}}\right)^{2} dx = \frac{x^{2}}{2} - 2x + \log x + C, \qquad x > 0.\]

EXERCISE 7.1 11–22 (part 2 of 27)

Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:
Exercise 11
$\displaystyle \int \frac{x^{3}+5 x^{2}-4}{x^{2}} d x$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2}+5 x+\frac{4}{x}+C$
Divide each term of the numerator by $\displaystyle x^{2}$ first ($\displaystyle x \neq 0$): \[\frac{x^{3} + 5x^{2} - 4}{x^{2}} = x + 5 - \frac{4}{x^{2}} = x + 5 - 4x^{-2}.\] Integrate term by term with the power rule; note $\displaystyle \int x^{-2}dx = \frac{x^{-1}}{-1} = -\frac{1}{x}$, so the last term changes sign: \[\int \left(x + 5 - 4x^{-2}\right) dx = \frac{x^{2}}{2} + 5x - 4\left(-\frac{1}{x}\right) + C.\] Hence \[\int \frac{x^{3} + 5x^{2} - 4}{x^{2}}\, dx = \frac{x^{2}}{2} + 5x + \frac{4}{x} + C, \qquad x \neq 0.\]
Exercise 12
$\displaystyle \int \frac{x^{3}+3 x+4}{\sqrt{x}} d x$
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NCERT’s answer
$\displaystyle \frac{2}{7} x^{\frac{7}{2}}+2 x^{\frac{3}{2}}+8 \sqrt{x}+C$
Write $\displaystyle \sqrt{x} = x^{1/2}$ and divide each term of the numerator by it, subtracting exponents ($\displaystyle x > 0$): \[\frac{x^{3} + 3x + 4}{\sqrt{x}} = x^{5/2} + 3x^{1/2} + 4x^{-1/2}.\] Apply the power rule $\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}$ to each term: \[\int x^{5/2}dx = \frac{x^{7/2}}{7/2} = \frac{2}{7}x^{7/2}, \quad 3\int x^{1/2}dx = 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}, \quad 4\int x^{-1/2}dx = 4\cdot\frac{x^{1/2}}{1/2} = 8x^{1/2}.\] Adding, \[\int \frac{x^{3} + 3x + 4}{\sqrt{x}}\, dx = \frac{2}{7}x^{7/2} + 2x^{3/2} + 8\sqrt{x} + C, \qquad x > 0.\]
Exercise 13
$\displaystyle \int \frac{x^{3}-x^{2}+x-1}{x-1} d x$
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NCERT’s answer
$\displaystyle \frac{x^{3}}{3}+x+\mathrm{C}$
Factorise the numerator by grouping so the denominator cancels: \[x^{3} - x^{2} + x - 1 = x^{2}(x - 1) + 1\,(x - 1) = (x - 1)\left(x^{2} + 1\right).\] Cancelling $\displaystyle x - 1$ is legitimate only where $\displaystyle x \neq 1$, which is exactly where the integrand is defined: \[\frac{x^{3} - x^{2} + x - 1}{x - 1} = x^{2} + 1, \qquad x \neq 1.\] Integrate by the power rule: \[\int \left(x^{2} + 1\right) dx = \frac{x^{3}}{3} + x + C.\] So \[\int \frac{x^{3} - x^{2} + x - 1}{x - 1}\, dx = \frac{x^{3}}{3} + x + C, \qquad x \neq 1.\]
Exercise 14
$\displaystyle \int(1-x) \sqrt{x} d x$
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NCERT’s answer
$\displaystyle \frac{2}{3} x^{\frac{3}{2}}-\frac{2}{5} x^{\frac{5}{2}}+C$
Multiply out first, since there is no product rule for integrals: \[(1 - x)\sqrt{x} = x^{1/2} - x^{3/2}.\] Apply the power rule to each term: \[\int x^{1/2}dx = \frac{x^{3/2}}{3/2} = \frac{2}{3}x^{3/2}, \qquad \int x^{3/2}dx = \frac{x^{5/2}}{5/2} = \frac{2}{5}x^{5/2}.\] Hence \[\int (1 - x)\sqrt{x}\, dx = \frac{2}{3}x^{3/2} - \frac{2}{5}x^{5/2} + C, \qquad x \geq 0.\]
Exercise 15
$\displaystyle \int \sqrt{x}\left(3 x^{2}+2 x+3\right) d x$
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NCERT’s answer
$\displaystyle \frac{6}{7} x^{\frac{7}{2}}+\frac{4}{5} x^{\frac{5}{2}}+2 x^{\frac{3}{2}}+\mathrm{C}$
Distribute $\displaystyle \sqrt{x} = x^{1/2}$ over the bracket, adding exponents: \[\sqrt{x}\left(3x^{2} + 2x + 3\right) = 3x^{5/2} + 2x^{3/2} + 3x^{1/2}.\] Integrate term by term with the power rule: \[3\cdot\frac{x^{7/2}}{7/2} = \frac{6}{7}x^{7/2}, \qquad 2\cdot\frac{x^{5/2}}{5/2} = \frac{4}{5}x^{5/2}, \qquad 3\cdot\frac{x^{3/2}}{3/2} = 2x^{3/2}.\] So \[\int \sqrt{x}\left(3x^{2} + 2x + 3\right) dx = \frac{6}{7}x^{7/2} + \frac{4}{5}x^{5/2} + 2x^{3/2} + C, \qquad x \geq 0.\]
Exercise 16
$\displaystyle \int\left(2 x-3 \cos x+e^{x}\right) d x$
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NCERT’s answer
$\displaystyle x^{2}-3 \sin x+e^{x}+\mathrm{C}$
Integrate term by term, using the power rule, $\displaystyle \int \cos x\, dx = \sin x$ and $\displaystyle \int e^{x}dx = e^{x}$: \[\int \left(2x - 3\cos x + e^{x}\right) dx = 2\cdot\frac{x^{2}}{2} - 3\sin x + e^{x} + C.\] Hence \[\int \left(2x - 3\cos x + e^{x}\right) dx = x^{2} - 3\sin x + e^{x} + C.\]
Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:
Exercise 17
$\displaystyle \int\left(2 x^{2}-3 \sin x+5 \sqrt{x}\right) d x$
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NCERT’s answer
$\displaystyle \frac{2}{3} x^{3}+3 \cos x+\frac{10}{3} x^{\frac{3}{2}}+\mathrm{C}$
Integrate term by term. The one to watch is the sine: $\displaystyle \int \sin x\, dx = -\cos x$, so the $\displaystyle -3\sin x$ term integrates to $\displaystyle +3\cos x$. \[2\int x^{2}dx = \frac{2}{3}x^{3}, \qquad -3\int \sin x\, dx = 3\cos x, \qquad 5\int x^{1/2}dx = 5\cdot\frac{x^{3/2}}{3/2} = \frac{10}{3}x^{3/2}.\] Hence \[\int \left(2x^{2} - 3\sin x + 5\sqrt{x}\right) dx = \frac{2}{3}x^{3} + 3\cos x + \frac{10}{3}x^{3/2} + C, \qquad x \geq 0.\]
Exercise 18
$\displaystyle \int \sec x(\sec x+\tan x) d x$
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NCERT’s answer
$\displaystyle \tan x+\sec x+\mathrm{C}$
Expand the product so that each piece is a standard derivative: \[\sec x\,(\sec x + \tan x) = \sec^{2}x + \sec x \tan x.\] Now use the standard results $\displaystyle \frac{d}{dx}(\tan x) = \sec^{2}x$ and $\displaystyle \frac{d}{dx}(\sec x) = \sec x \tan x$, i.e. \[\int \sec^{2}x\, dx = \tan x, \qquad \int \sec x \tan x\, dx = \sec x.\] Therefore \[\int \sec x\,(\sec x + \tan x)\, dx = \tan x + \sec x + C,\] valid on any interval on which $\displaystyle \cos x \neq 0$.
Find the following integrals in Exercises $\displaystyle 6$ to $\displaystyle 20$:
Exercise 19
$\displaystyle \int \frac{\sec ^{2} x}{\operatorname{cosec}^{2} x} d x$
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NCERT’s answer
$\displaystyle \tan x-x+\mathrm{C}$
Convert to sines and cosines before integrating: \[\frac{\sec^{2}x}{\mathrm{cosec}^{2}x} = \frac{1/\cos^{2}x}{1/\sin^{2}x} = \frac{\sin^{2}x}{\cos^{2}x} = \tan^{2}x.\] There is no direct formula for $\displaystyle \int \tan^{2}x\,dx$, so use the identity $\displaystyle 1 + \tan^{2}x = \sec^{2}x$, i.e. $\displaystyle \tan^{2}x = \sec^{2}x - 1$: \[\int \tan^{2}x\, dx = \int \left(\sec^{2}x - 1\right) dx = \tan x - x + C.\] Hence \[\int \frac{\sec^{2}x}{\mathrm{cosec}^{2}x}\, dx = \tan x - x + C,\] valid on any interval on which $\displaystyle \cos x \neq 0$.
Exercise 20
$\displaystyle \int \frac{2-3 \sin x}{\cos^{2} x} d x$
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NCERT’s answer
$\displaystyle 2 \tan x-3 \sec x+\mathrm{C}$
Split the fraction over the common denominator $\displaystyle \cos^{2}x$: \[\frac{2 - 3\sin x}{\cos^{2}x} = \frac{2}{\cos^{2}x} - 3\cdot\frac{\sin x}{\cos^{2}x} = 2\sec^{2}x - 3\sec x \tan x,\] since $\displaystyle \dfrac{\sin x}{\cos^{2}x} = \dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x} = \sec x \tan x$. Using $\displaystyle \int \sec^{2}x\,dx = \tan x$ and $\displaystyle \int \sec x \tan x\,dx = \sec x$: \[\int \frac{2 - 3\sin x}{\cos^{2}x}\, dx = 2\tan x - 3\sec x + C,\] valid on any interval on which $\displaystyle \cos x \neq 0$.
Choose the correct answer in Exercises $\displaystyle 21$ and 22.
Exercise 21
The anti derivative of $\displaystyle \left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)$ equals (A) $\displaystyle \frac{1}{3} x^{\frac{1}{3}}+2 x^{\frac{1}{2}}+\mathrm{C}$ (B) $\displaystyle \frac{2}{3} x^{\frac{2}{3}}+\frac{1}{2} x^{2}+\mathrm{C}$ (C) $\displaystyle \frac{2}{3} x^{\frac{3}{2}}+2 x^{\frac{1}{2}}+\mathrm{C}$ (D) $\displaystyle \frac{3}{2} x^{\frac{3}{2}}+\frac{1}{2} x^{\frac{1}{2}}+\mathrm{C}$
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NCERT’s answer
C
An anti derivative of a function is any function whose derivative is that function, so compute the integral by the power rule $\displaystyle \int x^{n}dx = \frac{x^{n+1}}{n+1}$ with $\displaystyle n = \frac{1}{2}$ and $\displaystyle n = -\frac{1}{2}$: \[\int \left(\sqrt{x} + \frac{1}{\sqrt{x}}\right) dx = \int \left(x^{1/2} + x^{-1/2}\right) dx = \frac{x^{3/2}}{3/2} + \frac{x^{1/2}}{1/2} + C = \frac{2}{3}x^{3/2} + 2x^{1/2} + C.\] Check by differentiating: $\displaystyle \frac{d}{dx}\left(\frac{2}{3}x^{3/2} + 2\sqrt{x}\right) = x^{1/2} + x^{-1/2}$, as required. This matches option (C). \[\boxed{\text{(C)}\ \ \frac{2}{3}x^{3/2} + 2x^{1/2} + C}\]
Exercise 22
If $\displaystyle \frac{d}{d x} f(x)=4 x^{3}-\frac{3}{x^{4}}$ such that $\displaystyle f(2)=0$. Then $\displaystyle f(x)$ is (A) $\displaystyle x^{4}+\frac{1}{x^{3}}-\frac{129}{8}$ (B) $\displaystyle x^{3}+\frac{1}{x^{4}}+\frac{129}{8}$ (C) $\displaystyle x^{4}+\frac{1}{x^{3}}+\frac{129}{8}$ (D) $\displaystyle x^{3}+\frac{1}{x^{4}}-\frac{129}{8}$
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NCERT’s answer
A
$\displaystyle f$ is an anti derivative of the given derivative, so integrate, writing $\displaystyle \frac{3}{x^{4}} = 3x^{-4}$: \[f(x) = \int \left(4x^{3} - 3x^{-4}\right) dx = 4\cdot\frac{x^{4}}{4} - 3\cdot\frac{x^{-3}}{-3} + C = x^{4} + \frac{1}{x^{3}} + C.\] The sign is the trap: $\displaystyle \int x^{-4}dx = \frac{x^{-3}}{-3}$, and the minus in front of $\displaystyle 3x^{-4}$ cancels it, leaving $\displaystyle +\frac{1}{x^{3}}$. Now use the condition $\displaystyle f(2) = 0$ to fix $\displaystyle C$: \[2^{4} + \frac{1}{2^{3}} + C = 0 \quad \Longrightarrow \quad 16 + \frac{1}{8} + C = 0 \quad \Longrightarrow \quad C = -\frac{129}{8}.\] Therefore \[f(x) = x^{4} + \frac{1}{x^{3}} - \frac{129}{8},\] which is option (A).

EXERCISE 7.2 1–10 (part 3 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:
Exercise 1
$\displaystyle \frac{2 x}{1+x^{2}}$
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NCERT’s answer
$\displaystyle \log \left(1+x^{2}\right)+\mathrm{C}$
Use the method of substitution: if $\displaystyle t=g(x)$, then $\displaystyle g'(x)\,dx=dt$. Here the numerator is exactly the derivative of the denominator, so put \[t=1+x^{2}\quad\Rightarrow\quad dt=2x\,dx.\] Then \[\int\frac{2x}{1+x^{2}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Since $\displaystyle 1+x^{2}>0$ for every real $\displaystyle x$, the modulus may be dropped. Resubstituting $\displaystyle t=1+x^{2}$, \[\int\frac{2x}{1+x^{2}}\,dx=\log\left(1+x^{2}\right)+\mathrm{C}.\]
Exercise 2
$\displaystyle \frac{(\log x)^{2}}{x}$
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NCERT’s answer
$\displaystyle \frac{1}{3}(\log |x|)^{3}+\mathrm{C}$
The factor $\displaystyle \dfrac{1}{x}$ is the derivative of $\displaystyle \log x$, which invites the substitution \[t=\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{(\log x)^{2}}{x}\,dx=\int t^{2}\,dt=\frac{t^{3}}{3}+\mathrm{C}.\] Resubstituting $\displaystyle t=\log x$, \[\int\frac{(\log x)^{2}}{x}\,dx=\frac{(\log x)^{3}}{3}+\mathrm{C}.\]
Exercise 3
$\displaystyle \frac{1}{x+x \log x}$
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NCERT’s answer
$\displaystyle \log |1+\log x|+C$
First take $\displaystyle x$ common in the denominator — this is the step that makes the substitution visible: \[\frac{1}{x+x\log x}=\frac{1}{x\,(1+\log x)}.\] Put \[t=1+\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{dx}{x(1+\log x)}=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Hence \[\int\frac{1}{x+x\log x}\,dx=\log\left|1+\log x\right|+\mathrm{C},\qquad x>0,\ \log x\neq-1.\]
Exercise 4
$\displaystyle \sin x \sin (\cos x)$
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NCERT’s answer
$\displaystyle \cos (\cos x)+\mathrm{C}$
The outer function has argument $\displaystyle \cos x$, and $\displaystyle \sin x$ is (up to sign) its derivative. Put \[t=\cos x\quad\Rightarrow\quad dt=-\sin x\,dx,\ \text{ i.e. }\ \sin x\,dx=-dt.\] Watch the sign here — it is the usual slip. Then \[\int\sin x\,\sin(\cos x)\,dx=\int \sin t\,(-dt)=-\int\sin t\,dt=\cos t+\mathrm{C}.\] Resubstituting $\displaystyle t=\cos x$, \[\int\sin x\,\sin(\cos x)\,dx=\cos(\cos x)+\mathrm{C}.\]
Exercise 5
$\displaystyle \sin (a x+b) \cos (a x+b)$
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NCERT’s answer
$\displaystyle -\frac{1}{4 a} \cos 2(a x+b)+\mathrm{C}$
Use the identity $\displaystyle 2\sin\theta\cos\theta=\sin 2\theta$ with $\displaystyle \theta=ax+b$: \[\sin(ax+b)\cos(ax+b)=\frac{1}{2}\sin\bigl(2(ax+b)\bigr).\] Therefore, using $\displaystyle \int\sin(kx+c)\,dx=-\dfrac{\cos(kx+c)}{k}$ with $\displaystyle k=2a\ (a\neq0)$, \[\int\sin(ax+b)\cos(ax+b)\,dx=\frac{1}{2}\int\sin(2ax+2b)\,dx=\frac{1}{2}\left(-\frac{\cos(2ax+2b)}{2a}\right)+\mathrm{C}.\] Hence \[\int\sin(ax+b)\cos(ax+b)\,dx=-\frac{\cos 2(ax+b)}{4a}+\mathrm{C}.\] (The substitution $\displaystyle t=\sin(ax+b)$ gives $\displaystyle \dfrac{\sin^{2}(ax+b)}{2a}+\mathrm{C}$; the two answers differ only by the constant $\displaystyle \dfrac{1}{4a}$.)
Exercise 6
$\displaystyle \sqrt{a x+b}$
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NCERT’s answer
$\displaystyle \frac{2}{3 a}(a x+b)^{\frac{3}{2}}+\mathrm{C}$
Put \[t=ax+b\quad\Rightarrow\quad dt=a\,dx,\ \text{ i.e. }\ dx=\frac{dt}{a}\qquad(a\neq0).\] Then, by the power rule $\displaystyle \int t^{n}\,dt=\dfrac{t^{n+1}}{n+1}$ with $\displaystyle n=\tfrac12$, \[\int\sqrt{ax+b}\,dx=\frac{1}{a}\int t^{1/2}\,dt=\frac{1}{a}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int\sqrt{ax+b}\,dx=\frac{2}{3a}(ax+b)^{3/2}+\mathrm{C}.\]
Exercise 7
$\displaystyle x \sqrt{x+2}$
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NCERT’s answer
$\displaystyle \frac{2}{5}(x+2)^{\frac{5}{2}}-\frac{4}{3}(x+2)^{\frac{3}{2}}+\mathrm{C}$
The factor $\displaystyle x$ outside is not the derivative of $\displaystyle x+2$, so substitute for the bracket and re-express $\displaystyle x$ in the new variable — that back-substitution is the step to get right. Put \[t=x+2\quad\Rightarrow\quad x=t-2,\quad dx=dt.\] Then \[\int x\sqrt{x+2}\,dx=\int (t-2)\,t^{1/2}\,dt=\int\left(t^{3/2}-2t^{1/2}\right)dt=\frac{2}{5}t^{5/2}-\frac{4}{3}t^{3/2}+\mathrm{C}.\] Resubstituting $\displaystyle t=x+2$, \[\int x\sqrt{x+2}\,dx=\frac{2}{5}(x+2)^{5/2}-\frac{4}{3}(x+2)^{3/2}+\mathrm{C}.\]
Exercise 8
$\displaystyle x \sqrt{1+2 x^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{6}\left(1+2 x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$
Here $\displaystyle x\,dx$ is proportional to the differential of $\displaystyle 1+2x^{2}$. Put \[t=1+2x^{2}\quad\Rightarrow\quad dt=4x\,dx,\ \text{ i.e. }\ x\,dx=\frac{dt}{4}.\] Then \[\int x\sqrt{1+2x^{2}}\,dx=\frac{1}{4}\int t^{1/2}\,dt=\frac{1}{4}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}=\frac{1}{6}t^{3/2}+\mathrm{C}.\] Hence \[\int x\sqrt{1+2x^{2}}\,dx=\frac{1}{6}\left(1+2x^{2}\right)^{3/2}+\mathrm{C}.\]
Exercise 9
$\displaystyle (4 x+2) \sqrt{x^{2}+x+1}$
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NCERT’s answer
$\displaystyle \frac{4}{3}\left(x^{2}+x+1\right)^{\frac{3}{2}}+\mathrm{C}$
Notice that $\displaystyle 4x+2=2(2x+1)$ and $\displaystyle 2x+1$ is the derivative of $\displaystyle x^{2}+x+1$. Put \[t=x^{2}+x+1\quad\Rightarrow\quad dt=(2x+1)\,dx,\ \text{ so }\ (4x+2)\,dx=2\,dt.\] Then \[\int(4x+2)\sqrt{x^{2}+x+1}\,dx=2\int t^{1/2}\,dt=2\cdot\frac{2}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int(4x+2)\sqrt{x^{2}+x+1}\,dx=\frac{4}{3}\left(x^{2}+x+1\right)^{3/2}+\mathrm{C}.\]
Exercise 10
$\displaystyle \frac{1}{x-\sqrt{x}}$
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NCERT’s answer
$\displaystyle 2 \log |\sqrt{x}-1|+\mathrm{C}$
A square root of $\displaystyle x$ with $\displaystyle x$ itself present: substitute for the root. Put \[t=\sqrt{x}\quad\Rightarrow\quad x=t^{2},\quad dx=2t\,dt\qquad(x>0).\] Then $\displaystyle x-\sqrt{x}=t^{2}-t=t(t-1)$, and the factor $\displaystyle t$ cancels: \[\int\frac{dx}{x-\sqrt{x}}=\int\frac{2t\,dt}{t(t-1)}=2\int\frac{dt}{t-1}=2\log|t-1|+\mathrm{C}.\] Resubstituting $\displaystyle t=\sqrt{x}$, \[\int\frac{1}{x-\sqrt{x}}\,dx=2\log\left|\sqrt{x}-1\right|+\mathrm{C},\qquad x>0,\ x\neq1.\]

EXERCISE 7.2 11–20 (part 4 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:
Exercise 11
$\displaystyle \frac{x}{\sqrt{x+4}}, x>0$
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NCERT’s answer
$\displaystyle \frac{2}{3} \sqrt{x+4}(x-8)+\mathrm{C}$
Put \[t=x+4\quad\Rightarrow\quad x=t-4,\quad dx=dt,\] so that (with $\displaystyle x>0$, hence $\displaystyle t>4>0$) \[\int\frac{x}{\sqrt{x+4}}\,dx=\int\frac{t-4}{\sqrt{t}}\,dt=\int\left(t^{1/2}-4t^{-1/2}\right)dt=\frac{2}{3}t^{3/2}-8t^{1/2}+\mathrm{C}.\] Resubstituting $\displaystyle t=x+4$, \[\int\frac{x}{\sqrt{x+4}}\,dx=\frac{2}{3}(x+4)^{3/2}-8\sqrt{x+4}+\mathrm{C}=\frac{2}{3}\sqrt{x+4}\,(x-8)+\mathrm{C}.\]
Exercise 12
$\displaystyle \left(x^{3}-1\right)^{\frac{1}{3}} x^{5}$
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NCERT’s answer
$\displaystyle \frac{1}{7}\left(x^{3}-1\right)^{\frac{7}{3}}+\frac{1}{4}\left(x^{3}-1\right)^{\frac{4}{3}}+\mathrm{C}$
Split the power of $\displaystyle x$ so that one part supplies the differential: $\displaystyle x^{5}\,dx=x^{3}\cdot x^{2}\,dx$. Put \[t=x^{3}-1\quad\Rightarrow\quad dt=3x^{2}\,dx,\quad x^{3}=t+1.\] Then \[\int\left(x^{3}-1\right)^{1/3}x^{5}\,dx=\frac{1}{3}\int t^{1/3}(t+1)\,dt=\frac{1}{3}\int\left(t^{4/3}+t^{1/3}\right)dt.\] Integrating, \[=\frac{1}{3}\left(\frac{3}{7}t^{7/3}+\frac{3}{4}t^{4/3}\right)+\mathrm{C}=\frac{1}{7}t^{7/3}+\frac{1}{4}t^{4/3}+\mathrm{C}.\] Hence \[\int\left(x^{3}-1\right)^{1/3}x^{5}\,dx=\frac{1}{7}\left(x^{3}-1\right)^{7/3}+\frac{1}{4}\left(x^{3}-1\right)^{4/3}+\mathrm{C}.\]
Exercise 13
$\displaystyle \frac{x^{2}}{\left(2+3 x^{3}\right)^{3}}$
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NCERT’s answer
$\displaystyle -\frac{1}{18\left(2+3 x^{3}\right)^{2}}+\mathrm{C}$
Put \[t=2+3x^{3}\quad\Rightarrow\quad dt=9x^{2}\,dx,\ \text{ i.e. }\ x^{2}\,dx=\frac{dt}{9}.\] Then \[\int\frac{x^{2}}{\left(2+3x^{3}\right)^{3}}\,dx=\frac{1}{9}\int t^{-3}\,dt=\frac{1}{9}\cdot\frac{t^{-2}}{-2}+\mathrm{C}=-\frac{1}{18t^{2}}+\mathrm{C}.\] Hence \[\int\frac{x^{2}}{\left(2+3x^{3}\right)^{3}}\,dx=-\frac{1}{18\left(2+3x^{3}\right)^{2}}+\mathrm{C}.\]
Exercise 14
$\displaystyle \frac{1}{x(\log x)^{m}}, x>0, m \neq 1$
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NCERT’s answer
$\displaystyle \frac{(\log x)^{1-m}}{1-m}+\mathrm{C}$
Put \[t=\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{dx}{x(\log x)^{m}}=\int t^{-m}\,dt.\] The power rule $\displaystyle \int t^{n}\,dt=\dfrac{t^{n+1}}{n+1}$ applies because $\displaystyle -m\neq-1$, which is exactly why the question stipulates $\displaystyle m\neq1$: \[\int t^{-m}\,dt=\frac{t^{-m+1}}{1-m}+\mathrm{C}.\] Hence \[\int\frac{1}{x(\log x)^{m}}\,dx=\frac{(\log x)^{1-m}}{1-m}+\mathrm{C},\qquad x>0,\ m\neq1.\]
Exercise 15
$\displaystyle \frac{x}{9-4 x^{2}}$
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NCERT’s answer
$\displaystyle -\frac{1}{8} \log \left|9-4 \mathrm{x}^{2}\right|+\mathrm{C}$
The numerator is a constant multiple of the derivative of the denominator. Put \[t=9-4x^{2}\quad\Rightarrow\quad dt=-8x\,dx,\ \text{ i.e. }\ x\,dx=-\frac{dt}{8}.\] Then \[\int\frac{x}{9-4x^{2}}\,dx=-\frac{1}{8}\int\frac{dt}{t}=-\frac{1}{8}\log|t|+\mathrm{C}.\] Since $\displaystyle 9-4x^{2}$ can be negative, the modulus must be kept: \[\int\frac{x}{9-4x^{2}}\,dx=-\frac{1}{8}\log\left|9-4x^{2}\right|+\mathrm{C}.\]
Exercise 16
$\displaystyle e^{2 x+3}$
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NCERT’s answer
$\displaystyle \frac{1}{2} e^{2 x+3}+\mathrm{C}$
Put \[t=2x+3\quad\Rightarrow\quad dt=2\,dx,\ \text{ i.e. }\ dx=\frac{dt}{2}.\] Using $\displaystyle \int e^{t}\,dt=e^{t}+\mathrm{C}$, \[\int e^{2x+3}\,dx=\frac{1}{2}\int e^{t}\,dt=\frac{1}{2}e^{t}+\mathrm{C}.\] Hence \[\int e^{2x+3}\,dx=\frac{1}{2}e^{2x+3}+\mathrm{C}.\]
Exercise 17
$\displaystyle \frac{x}{e^{x^{2}}}$
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NCERT’s answer
$\displaystyle -\frac{1}{2 e^{x^{2}}}+\mathrm{C}$
Write the integrand as $\displaystyle \dfrac{x}{e^{x^{2}}}=x\,e^{-x^{2}}$ and put \[t=x^{2}\quad\Rightarrow\quad dt=2x\,dx,\ \text{ i.e. }\ x\,dx=\frac{dt}{2}.\] Then \[\int x\,e^{-x^{2}}\,dx=\frac{1}{2}\int e^{-t}\,dt=\frac{1}{2}\left(-e^{-t}\right)+\mathrm{C}=-\frac{1}{2}e^{-t}+\mathrm{C}.\] Hence \[\int\frac{x}{e^{x^{2}}}\,dx=-\frac{1}{2e^{x^{2}}}+\mathrm{C}.\]
Exercise 18
$\displaystyle \frac{e^{\tan ^{-1} x}}{1+x^{2}}$
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NCERT’s answer
$\displaystyle e^{\tan ^{-1} x}+\mathrm{C}$
Recall $\displaystyle \dfrac{d}{dx}\left(\tan^{-1}x\right)=\dfrac{1}{1+x^{2}}$, so the factor $\displaystyle \dfrac{dx}{1+x^{2}}$ is exactly the differential of the exponent. Put \[t=\tan^{-1}x\quad\Rightarrow\quad dt=\frac{dx}{1+x^{2}}.\] Then \[\int\frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx=\int e^{t}\,dt=e^{t}+\mathrm{C}.\] Hence \[\int\frac{e^{\tan^{-1}x}}{1+x^{2}}\,dx=e^{\tan^{-1}x}+\mathrm{C}.\]
Exercise 19
$\displaystyle \frac{e^{2 x}-1}{e^{2 x}+1}$
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NCERT’s answer
$\displaystyle \log \left(e^{x}+e^{-x}\right)+\mathrm{C}$
Divide numerator and denominator by $\displaystyle e^{x}$ — this is the step that turns the integrand into a derivative-over-function form: \[\frac{e^{2x}-1}{e^{2x}+1}=\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}.\] Put \[t=e^{x}+e^{-x}\quad\Rightarrow\quad dt=\left(e^{x}-e^{-x}\right)dx.\] Then \[\int\frac{e^{2x}-1}{e^{2x}+1}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Since $\displaystyle e^{x}+e^{-x}>0$ always, \[\int\frac{e^{2x}-1}{e^{2x}+1}\,dx=\log\left(e^{x}+e^{-x}\right)+\mathrm{C}.\]
Exercise 20
$\displaystyle \frac{e^{2 x}-e^{-2 x}}{e^{2 x}+e^{-2 x}}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left(e^{2 x}+e^{-2 x}\right)+\mathrm{C}$
The numerator is, up to a factor $\displaystyle 2$, the derivative of the denominator: \[\frac{d}{dx}\left(e^{2x}+e^{-2x}\right)=2\left(e^{2x}-e^{-2x}\right).\] Put \[t=e^{2x}+e^{-2x}\quad\Rightarrow\quad dt=2\left(e^{2x}-e^{-2x}\right)dx.\] Then \[\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\log|t|+\mathrm{C}.\] Since $\displaystyle e^{2x}+e^{-2x}>0$, \[\int\frac{e^{2x}-e^{-2x}}{e^{2x}+e^{-2x}}\,dx=\frac{1}{2}\log\left(e^{2x}+e^{-2x}\right)+\mathrm{C}.\]

EXERCISE 7.2 21–30 (part 5 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:
Exercise 21
$\displaystyle \tan ^{2}(2 x-3)$
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NCERT’s answer
$\displaystyle \frac{1}{2} \tan (2 x-3)-x+C$
There is no direct formula for $\displaystyle \int\tan^{2}\theta\,d\theta$; convert it first using the identity $\displaystyle \sec^{2}\theta=1+\tan^{2}\theta$, i.e. $\displaystyle \tan^{2}\theta=\sec^{2}\theta-1$, with $\displaystyle \theta=2x-3$: \[\int\tan^{2}(2x-3)\,dx=\int\left[\sec^{2}(2x-3)-1\right]dx.\] With $\displaystyle t=2x-3$, $\displaystyle dt=2\,dx$, and $\displaystyle \int\sec^{2}t\,dt=\tan t$, \[=\frac{1}{2}\tan(2x-3)-x+\mathrm{C}.\] Hence \[\int\tan^{2}(2x-3)\,dx=\frac{1}{2}\tan(2x-3)-x+\mathrm{C}.\]
Exercise 22
$\displaystyle \sec ^{2}(7-4 x)$
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NCERT’s answer
$\displaystyle -\frac{1}{4} \tan (7-4 x)+\mathrm{C}$
Put \[t=7-4x\quad\Rightarrow\quad dt=-4\,dx,\ \text{ i.e. }\ dx=-\frac{dt}{4}.\] The negative sign from the coefficient of $\displaystyle x$ is the usual slip here. Using $\displaystyle \int\sec^{2}t\,dt=\tan t+\mathrm{C}$, \[\int\sec^{2}(7-4x)\,dx=-\frac{1}{4}\int\sec^{2}t\,dt=-\frac{1}{4}\tan t+\mathrm{C}.\] Hence \[\int\sec^{2}(7-4x)\,dx=-\frac{1}{4}\tan(7-4x)+\mathrm{C}.\]
Exercise 23
$\displaystyle \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}}$
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NCERT’s answer
$\displaystyle \frac{1}{2}\left(\sin ^{-1} x\right)^{2}+\mathrm{C}$
Recall $\displaystyle \dfrac{d}{dx}\left(\sin^{-1}x\right)=\dfrac{1}{\sqrt{1-x^{2}}}$, so put \[t=\sin^{-1}x\quad\Rightarrow\quad dt=\frac{dx}{\sqrt{1-x^{2}}}\qquad(-1<x<1).\] Then \[\int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\,dx=\int t\,dt=\frac{t^{2}}{2}+\mathrm{C}.\] Hence \[\int\frac{\sin^{-1}x}{\sqrt{1-x^{2}}}\,dx=\frac{\left(\sin^{-1}x\right)^{2}}{2}+\mathrm{C},\qquad -1<x<1.\]
Exercise 24
$\displaystyle \frac{2 \cos x-3 \sin x}{6 \cos x+4 \sin x}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log |2 \sin x+3 \cos x|+C$
Differentiate the denominator and compare with the numerator: \[\frac{d}{dx}\left(6\cos x+4\sin x\right)=-6\sin x+4\cos x=2\left(2\cos x-3\sin x\right).\] So the numerator is exactly half the derivative of the denominator. Put \[t=6\cos x+4\sin x\quad\Rightarrow\quad dt=2(2\cos x-3\sin x)\,dx.\] Then \[\int\frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx=\frac{1}{2}\int\frac{dt}{t}=\frac{1}{2}\log|t|+\mathrm{C}.\] Hence \[\int\frac{2\cos x-3\sin x}{6\cos x+4\sin x}\,dx=\frac{1}{2}\log\left|6\cos x+4\sin x\right|+\mathrm{C},\] which (taking $\displaystyle 2$ out of the logarithm into the constant) may also be written as $\displaystyle \dfrac{1}{2}\log\left|3\cos x+2\sin x\right|+\mathrm{C}$.
Exercise 25
$\displaystyle \frac{1}{\cos ^{2} x(1-\tan x)^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{(1-\tan x)}+\mathrm{C}$
Write $\displaystyle \dfrac{1}{\cos^{2}x}=\sec^{2}x$, so that \[\int\frac{dx}{\cos^{2}x(1-\tan x)^{2}}=\int\frac{\sec^{2}x}{(1-\tan x)^{2}}\,dx.\] Put \[t=1-\tan x\quad\Rightarrow\quad dt=-\sec^{2}x\,dx,\ \text{ i.e. }\ \sec^{2}x\,dx=-dt.\] Then \[\int\frac{\sec^{2}x}{(1-\tan x)^{2}}\,dx=-\int\frac{dt}{t^{2}}=-\left(-\frac{1}{t}\right)+\mathrm{C}=\frac{1}{t}+\mathrm{C}.\] Hence \[\int\frac{1}{\cos^{2}x(1-\tan x)^{2}}\,dx=\frac{1}{1-\tan x}+\mathrm{C}.\]
Exercise 26
$\displaystyle \frac{\cos \sqrt{x}}{\sqrt{x}}$
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NCERT’s answer
$\displaystyle 2 \sin \sqrt{x}+\mathrm{C}$
Put \[t=\sqrt{x}\quad\Rightarrow\quad dt=\frac{1}{2\sqrt{x}}\,dx,\ \text{ i.e. }\ \frac{dx}{\sqrt{x}}=2\,dt\qquad(x>0).\] Then \[\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\int\cos t\,dt=2\sin t+\mathrm{C}.\] Hence \[\int\frac{\cos\sqrt{x}}{\sqrt{x}}\,dx=2\sin\sqrt{x}+\mathrm{C},\qquad x>0.\]
Exercise 27
$\displaystyle \sqrt{\sin 2 x} \cos 2 x$
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NCERT’s answer
$\displaystyle \frac{1}{3}(\sin 2 x)^{\frac{3}{2}}+\mathrm{C}$
Here $\displaystyle \cos 2x\,dx$ is proportional to the differential of $\displaystyle \sin 2x$. Put \[t=\sin 2x\quad\Rightarrow\quad dt=2\cos 2x\,dx,\ \text{ i.e. }\ \cos 2x\,dx=\frac{dt}{2}.\] Then \[\int\sqrt{\sin 2x}\,\cos 2x\,dx=\frac{1}{2}\int t^{1/2}\,dt=\frac{1}{2}\cdot\frac{2}{3}t^{3/2}+\mathrm{C}=\frac{1}{3}t^{3/2}+\mathrm{C}.\] Hence \[\int\sqrt{\sin 2x}\,\cos 2x\,dx=\frac{1}{3}\left(\sin 2x\right)^{3/2}+\mathrm{C},\] valid on any interval where $\displaystyle \sin 2x\geq0$ (so that the square root is defined).
Exercise 28
$\displaystyle \frac{\cos x}{\sqrt{1+\sin x}}$
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NCERT’s answer
$\displaystyle 2 \sqrt{1+\sin x}+\mathrm{C}$
Put \[t=1+\sin x\quad\Rightarrow\quad dt=\cos x\,dx.\] Then \[\int\frac{\cos x}{\sqrt{1+\sin x}}\,dx=\int\frac{dt}{\sqrt{t}}=\int t^{-1/2}\,dt=2t^{1/2}+\mathrm{C}.\] Hence \[\int\frac{\cos x}{\sqrt{1+\sin x}}\,dx=2\sqrt{1+\sin x}+\mathrm{C},\] taken on an interval where $\displaystyle 1+\sin x>0$.
Exercise 29
$\displaystyle \cot x \log \sin x$
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NCERT’s answer
$\displaystyle \frac{1}{2}(\log \sin x)^{2}+\mathrm{C}$
Note that $\displaystyle \dfrac{d}{dx}\left(\log\sin x\right)=\dfrac{\cos x}{\sin x}=\cot x$, so the whole factor $\displaystyle \cot x\,dx$ is the differential of the logarithm. Put \[t=\log\sin x\quad\Rightarrow\quad dt=\cot x\,dx\qquad(\sin x>0).\] Then \[\int\cot x\,\log\sin x\,dx=\int t\,dt=\frac{t^{2}}{2}+\mathrm{C}.\] Hence \[\int\cot x\,\log\sin x\,dx=\frac{\left(\log\sin x\right)^{2}}{2}+\mathrm{C},\qquad \sin x>0.\]
Exercise 30
$\displaystyle \frac{\sin x}{1+\cos x}$
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NCERT’s answer
$\displaystyle -\log |1+\cos x|+\mathrm{C}$
Put \[t=1+\cos x\quad\Rightarrow\quad dt=-\sin x\,dx,\ \text{ i.e. }\ \sin x\,dx=-dt.\] Then \[\int\frac{\sin x}{1+\cos x}\,dx=-\int\frac{dt}{t}=-\log|t|+\mathrm{C}.\] Hence \[\int\frac{\sin x}{1+\cos x}\,dx=-\log\left|1+\cos x\right|+\mathrm{C},\qquad \cos x\neq-1.\]

EXERCISE 7.2 31–39 (part 6 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 37$:
Exercise 31
$\displaystyle \frac{\sin x}{(1+\cos x)^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{1+\cos x}+C$
Use the same substitution as before, now with the square in the denominator. Put \[t=1+\cos x\quad\Rightarrow\quad dt=-\sin x\,dx,\ \text{ i.e. }\ \sin x\,dx=-dt.\] Then \[\int\frac{\sin x}{(1+\cos x)^{2}}\,dx=-\int\frac{dt}{t^{2}}=-\left(-\frac{1}{t}\right)+\mathrm{C}=\frac{1}{t}+\mathrm{C}.\] Hence \[\int\frac{\sin x}{(1+\cos x)^{2}}\,dx=\frac{1}{1+\cos x}+\mathrm{C},\qquad \cos x\neq-1.\]
Exercise 32
$\displaystyle \frac{1}{1+\cot x}$
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NCERT’s answer
$\displaystyle \frac{x}{2}-\frac{1}{2} \log |\cos x+\sin x|+\mathrm{C}$
Convert to sine and cosine: \[\frac{1}{1+\cot x}=\frac{1}{1+\dfrac{\cos x}{\sin x}}=\frac{\sin x}{\sin x+\cos x}.\] Now split the numerator into a part equal to the denominator and a part equal to its derivative — this is the standard device for $\displaystyle \dfrac{a\sin x+b\cos x}{c\sin x+d\cos x}$: \[\sin x=\frac{1}{2}\Bigl[(\sin x+\cos x)+(\sin x-\cos x)\Bigr].\] Therefore \[\int\frac{\sin x}{\sin x+\cos x}\,dx=\frac{1}{2}\int dx+\frac{1}{2}\int\frac{\sin x-\cos x}{\sin x+\cos x}\,dx.\] For the second integral put $\displaystyle u=\sin x+\cos x$, so $\displaystyle du=(\cos x-\sin x)\,dx=-(\sin x-\cos x)\,dx$, giving \[\frac{1}{2}\int\frac{\sin x-\cos x}{\sin x+\cos x}\,dx=-\frac{1}{2}\int\frac{du}{u}=-\frac{1}{2}\log|u|+\mathrm{C}.\] Hence \[\int\frac{1}{1+\cot x}\,dx=\frac{x}{2}-\frac{1}{2}\log\left|\sin x+\cos x\right|+\mathrm{C}.\]
Exercise 33
$\displaystyle \frac{1}{1-\tan x}$
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NCERT’s answer
$\displaystyle \frac{x}{2}-\frac{1}{2} \log |\cos x-\sin x|+C$
Convert to sine and cosine: \[\frac{1}{1-\tan x}=\frac{1}{1-\dfrac{\sin x}{\cos x}}=\frac{\cos x}{\cos x-\sin x}.\] Split the numerator as (denominator) plus (derivative-type term): \[\cos x=\frac{1}{2}\Bigl[(\cos x-\sin x)+(\cos x+\sin x)\Bigr].\] Therefore \[\int\frac{\cos x}{\cos x-\sin x}\,dx=\frac{1}{2}\int dx+\frac{1}{2}\int\frac{\cos x+\sin x}{\cos x-\sin x}\,dx.\] For the second integral put $\displaystyle u=\cos x-\sin x$, so $\displaystyle du=-(\sin x+\cos x)\,dx$; note the sign: \[\frac{1}{2}\int\frac{\cos x+\sin x}{\cos x-\sin x}\,dx=-\frac{1}{2}\int\frac{du}{u}=-\frac{1}{2}\log|u|+\mathrm{C}.\] Hence \[\int\frac{1}{1-\tan x}\,dx=\frac{x}{2}-\frac{1}{2}\log\left|\cos x-\sin x\right|+\mathrm{C}.\]
Exercise 34
$\displaystyle \frac{\sqrt{\tan x}}{\sin x \cos x}$
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NCERT’s answer
$\displaystyle 2 \sqrt{\tan x}+\mathrm{C}$
Rewrite the denominator so that $\displaystyle \sec^{2}x$ appears — that is the key manipulation: \[\sin x\cos x=\cos^{2}x\cdot\frac{\sin x}{\cos x}=\cos^{2}x\,\tan x.\] Hence \[\frac{\sqrt{\tan x}}{\sin x\cos x}=\frac{\sqrt{\tan x}\,\sec^{2}x}{\tan x}=\frac{\sec^{2}x}{\sqrt{\tan x}}.\] Put \[t=\tan x\quad\Rightarrow\quad dt=\sec^{2}x\,dx.\] Then \[\int\frac{\sec^{2}x}{\sqrt{\tan x}}\,dx=\int t^{-1/2}\,dt=2t^{1/2}+\mathrm{C}.\] Hence \[\int\frac{\sqrt{\tan x}}{\sin x\cos x}\,dx=2\sqrt{\tan x}+\mathrm{C},\] on an interval where $\displaystyle \tan x>0$.
Exercise 35
$\displaystyle \frac{(1+\log x)^{2}}{x}$
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NCERT’s answer
$\displaystyle \frac{1}{3}(1+\log x)^{3}+\mathrm{C}$
Put \[t=1+\log x\quad\Rightarrow\quad dt=\frac{1}{x}\,dx\qquad(x>0).\] Then \[\int\frac{(1+\log x)^{2}}{x}\,dx=\int t^{2}\,dt=\frac{t^{3}}{3}+\mathrm{C}.\] Hence \[\int\frac{(1+\log x)^{2}}{x}\,dx=\frac{(1+\log x)^{3}}{3}+\mathrm{C},\qquad x>0.\]
Exercise 36
$\displaystyle \frac{(x+1)(x+\log x)^{2}}{x}$
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NCERT’s answer
$\displaystyle \frac{1}{3}(x+\log x)^{3}+\mathrm{C}$
Regroup the integrand so that the loose factor becomes a derivative: \[\frac{(x+1)(x+\log x)^{2}}{x}=\left(1+\frac{1}{x}\right)(x+\log x)^{2},\] and observe that $\displaystyle \dfrac{d}{dx}(x+\log x)=1+\dfrac{1}{x}$. Put \[t=x+\log x\quad\Rightarrow\quad dt=\left(1+\frac{1}{x}\right)dx\qquad(x>0).\] Then \[\int\frac{(x+1)(x+\log x)^{2}}{x}\,dx=\int t^{2}\,dt=\frac{t^{3}}{3}+\mathrm{C}.\] Hence \[\int\frac{(x+1)(x+\log x)^{2}}{x}\,dx=\frac{(x+\log x)^{3}}{3}+\mathrm{C},\qquad x>0.\]
Exercise 37
$\displaystyle \frac{x^{3} \sin \left(\tan ^{-1} x^{4}\right)}{1+x^{8}}$
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NCERT’s answer
$\displaystyle -\frac{1}{4} \cos \left(\tan ^{-1} x^{4}\right)+\mathrm{C}$
Note that $\displaystyle 1+x^{8}=1+\left(x^{4}\right)^{2}$, so \[\frac{d}{dx}\left(\tan^{-1}x^{4}\right)=\frac{4x^{3}}{1+x^{8}}.\] Put \[t=\tan^{-1}x^{4}\quad\Rightarrow\quad dt=\frac{4x^{3}}{1+x^{8}}\,dx,\ \text{ i.e. }\ \frac{x^{3}}{1+x^{8}}\,dx=\frac{dt}{4}.\] Then \[\int\frac{x^{3}\sin\left(\tan^{-1}x^{4}\right)}{1+x^{8}}\,dx=\frac{1}{4}\int\sin t\,dt=-\frac{1}{4}\cos t+\mathrm{C}.\] Hence \[\int\frac{x^{3}\sin\left(\tan^{-1}x^{4}\right)}{1+x^{8}}\,dx=-\frac{1}{4}\cos\left(\tan^{-1}x^{4}\right)+\mathrm{C}.\]
Choose the correct answer in Exercises $\displaystyle 38$ and 39.
Exercise 38
$\displaystyle \int \frac{10 x^{9}+10^{x} \log _{e} 10 d x}{x^{10}+10^{x}}$ equals (A) $\displaystyle 10^{x}-x^{10}+\mathrm{C}$ (B) $\displaystyle 10^{x}+x^{10}+\mathrm{C}$ (C) $\displaystyle \left(10^{x}-x^{10}\right)^{-1}+\mathrm{C}$ (D) $\displaystyle \log \left(10^{x}+x^{10}\right)+\mathrm{C}$
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NCERT’s answer
D
Test whether the numerator is the derivative of the denominator. With $\displaystyle f(x)=x^{10}+10^{x}$, and using $\displaystyle \dfrac{d}{dx}\left(a^{x}\right)=a^{x}\log_{e}a$, \[f'(x)=10x^{9}+10^{x}\log_{e}10,\] which is exactly the numerator. So the integral is of the form $\displaystyle \int\dfrac{f'(x)}{f(x)}\,dx$. Put $\displaystyle t=x^{10}+10^{x}$, $\displaystyle dt=\left(10x^{9}+10^{x}\log_{e}10\right)dx$: \[\int\frac{10x^{9}+10^{x}\log_{e}10}{x^{10}+10^{x}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}=\log\left(x^{10}+10^{x}\right)+\mathrm{C}.\] The correct option is (D) $\displaystyle \log\left(10^{x}+x^{10}\right)+\mathrm{C}$.
Exercise 39
$\displaystyle \int \frac{d x}{\sin ^{2} x \cos ^{2} x}$ equals (A) $\displaystyle \tan x+\cot x+\mathrm{C}$ (B) $\displaystyle \tan x-\cot x+\mathrm{C}$ (C) $\displaystyle \tan x \cot x+\mathrm{C}$ (D) $\displaystyle \tan x-\cot 2 x+\mathrm{C}$
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NCERT’s answer
B
Replace the $\displaystyle 1$ in the numerator using the identity $\displaystyle \sin^{2}x+\cos^{2}x=1$ — this is the step that splits the integral: \[\frac{1}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x+\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{1}{\cos^{2}x}+\frac{1}{\sin^{2}x}=\sec^{2}x+\mathrm{cosec}^{2}x.\] Therefore, using $\displaystyle \int\sec^{2}x\,dx=\tan x$ and $\displaystyle \int\mathrm{cosec}^{2}x\,dx=-\cot x$, \[\int\frac{dx}{\sin^{2}x\cos^{2}x}=\tan x-\cot x+\mathrm{C}.\] The correct option is (B) $\displaystyle \tan x-\cot x+\mathrm{C}$.

EXERCISE 7.3 1–10 (part 7 of 27)

Find the integrals of the functions in Exercises $\displaystyle 1$ to $\displaystyle 22$:
Exercise 1
$\displaystyle \sin ^{2}(2 x+5)$
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NCERT’s answer
$\displaystyle \frac{x}{2}-\frac{1}{8} \sin (4 x+10)+\mathrm{C}$
Apply the power-reducing identity $\displaystyle \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}$ with $\displaystyle \theta=2x+5$, so that $\displaystyle 2\theta=4x+10$: \[\sin^{2}(2x+5)=\frac{1-\cos(4x+10)}{2}.\] Now integrate, using $\displaystyle \int\cos(ax+b)\,dx=\dfrac{\sin(ax+b)}{a}$ (the factor $\displaystyle \tfrac1a$ is what the chain rule forces): \[\int\sin^{2}(2x+5)\,dx=\frac{1}{2}\int dx-\frac{1}{2}\int\cos(4x+10)\,dx=\frac{x}{2}-\frac{1}{2}\cdot\frac{\sin(4x+10)}{4}+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{x}{2}-\dfrac{1}{8}\sin(4x+10)+\mathrm{C}$.
Exercise 2
$\displaystyle \sin 3 x \cos 4 x$
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NCERT’s answer
$\displaystyle -\frac{1}{14} \cos 7 x+\frac{1}{2} \cos x+\mathrm{C}$
A product of sines and cosines has no direct antiderivative, so convert it to a sum by $\displaystyle \sin A\cos B=\tfrac12\big[\sin(A+B)+\sin(A-B)\big]$. With $\displaystyle A=3x,\;B=4x$: \[\sin 3x\cos 4x=\tfrac12\big[\sin 7x+\sin(-x)\big]=\tfrac12\big[\sin 7x-\sin x\big],\] using $\displaystyle \sin(-x)=-\sin x$. Integrating, \[\int\sin 3x\cos 4x\,dx=\frac{1}{2}\left[-\frac{\cos 7x}{7}+\cos x\right]+\mathrm{C}.\] Final answer: $\displaystyle -\dfrac{\cos 7x}{14}+\dfrac{\cos x}{2}+\mathrm{C}$.
Exercise 3
$\displaystyle \cos 2 x \cos 4 x \cos 6 x$
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NCERT’s answer
$\displaystyle \frac{1}{4}\left[\frac{1}{12} \sin 12 x+x+\frac{1}{8} \sin 8 x+\frac{1}{4} \sin 4 x\right]+C$
Reduce the triple product to sums in two stages, using $\displaystyle \cos A\cos B=\tfrac12\big[\cos(A+B)+\cos(A-B)\big]$. First pair the first two factors: \[\cos 2x\cos 4x=\tfrac12\big[\cos 6x+\cos 2x\big].\] Multiply by $\displaystyle \cos 6x$: \[\cos 2x\cos 4x\cos 6x=\tfrac12\big[\cos^{2}6x+\cos 2x\cos 6x\big]=\tfrac12\left[\frac{1+\cos 12x}{2}+\frac{\cos 8x+\cos 4x}{2}\right],\] where $\displaystyle \cos^{2}6x=\dfrac{1+\cos 12x}{2}$ and $\displaystyle \cos 2x\cos 6x=\tfrac12[\cos 8x+\cos 4x]$. Hence the integrand is \[\tfrac14\big[1+\cos 12x+\cos 8x+\cos 4x\big].\] Integrating term by term, \[\int\cos 2x\cos 4x\cos 6x\,dx=\frac{1}{4}\left[x+\frac{\sin 12x}{12}+\frac{\sin 8x}{8}+\frac{\sin 4x}{4}\right]+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{x}{4}+\dfrac{\sin 12x}{48}+\dfrac{\sin 8x}{32}+\dfrac{\sin 4x}{16}+\mathrm{C}$.
Exercise 4
$\displaystyle \sin ^{3}(2 x+1)$
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NCERT’s answer
$\displaystyle -\frac{1}{2} \cos (2 x+1)+\frac{1}{6} \cos ^{3}(2 x+1)+\mathrm{C}$
Use the triple-angle identity $\displaystyle \sin 3\theta=3\sin\theta-4\sin^{3}\theta$, i.e. $\displaystyle \sin^{3}\theta=\dfrac{3\sin\theta-\sin 3\theta}{4}$. Take $\displaystyle \theta=2x+1$, so $\displaystyle 3\theta=6x+3$: \[\sin^{3}(2x+1)=\frac{3\sin(2x+1)-\sin(6x+3)}{4}.\] Integrate, with $\displaystyle \int\sin(ax+b)\,dx=-\dfrac{\cos(ax+b)}{a}$: \[\int\sin^{3}(2x+1)\,dx=\frac{3}{4}\left(-\frac{\cos(2x+1)}{2}\right)-\frac{1}{4}\left(-\frac{\cos(6x+3)}{6}\right)+\mathrm{C}.\] Final answer: $\displaystyle -\dfrac{3}{8}\cos(2x+1)+\dfrac{1}{24}\cos(6x+3)+\mathrm{C}$. (Equivalently, splitting off one sine and putting $\displaystyle t=\cos(2x+1)$ gives $\displaystyle -\tfrac12\cos(2x+1)+\tfrac16\cos^{3}(2x+1)+\mathrm{C}$; the two answers differ only by a constant.)
Exercise 5
$\displaystyle \sin ^{3} x \cos ^{3} x$
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NCERT’s answer
$\displaystyle \frac{1}{6} \cos ^{6} x-\frac{1}{4} \cos ^{4} x+\mathrm{C}$
Pull the product into a single angle with $\displaystyle \sin x\cos x=\tfrac12\sin 2x$: \[\sin^{3}x\cos^{3}x=(\sin x\cos x)^{3}=\left(\frac{\sin 2x}{2}\right)^{3}=\frac{\sin^{3}2x}{8}.\] Now use $\displaystyle \sin^{3}\theta=\dfrac{3\sin\theta-\sin 3\theta}{4}$ with $\displaystyle \theta=2x$: \[\frac{\sin^{3}2x}{8}=\frac{3\sin 2x-\sin 6x}{32}.\] Integrating, \[\int\sin^{3}x\cos^{3}x\,dx=\frac{1}{32}\left[3\left(-\frac{\cos 2x}{2}\right)+\frac{\cos 6x}{6}\right]+\mathrm{C}.\] Final answer: $\displaystyle -\dfrac{3\cos 2x}{64}+\dfrac{\cos 6x}{192}+\mathrm{C}$. (The substitution $\displaystyle t=\cos x$ gives the equivalent form $\displaystyle -\dfrac{\cos^{4}x}{4}+\dfrac{\cos^{6}x}{6}+\mathrm{C}$.)
Exercise 6
$\displaystyle \sin x \sin 2 x \sin 3 x$
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NCERT’s answer
$\displaystyle \frac{1}{4}\left[\frac{1}{6} \cos 6 x-\frac{1}{4} \cos 4 x-\frac{1}{2} \cos 2 x\right]+\mathrm{C}$
Pair the two factors whose angles combine cleanly. By $\displaystyle \sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big]$, \[\sin x\sin 3x=\tfrac12\big[\cos 2x-\cos 4x\big].\] Multiply by $\displaystyle \sin 2x$ and use $\displaystyle \sin A\cos B=\tfrac12[\sin(A+B)+\sin(A-B)]$: \[\sin x\sin 2x\sin 3x=\tfrac12\big[\sin 2x\cos 2x-\sin 2x\cos 4x\big]=\tfrac12\left[\frac{\sin 4x}{2}-\frac{\sin 6x-\sin 2x}{2}\right],\] since $\displaystyle \sin 2x\cos 4x=\tfrac12[\sin 6x+\sin(-2x)]=\tfrac12[\sin 6x-\sin 2x]$. So the integrand is \[\tfrac14\big[\sin 4x-\sin 6x+\sin 2x\big].\] Integrating, \[\int\sin x\sin 2x\sin 3x\,dx=\frac{1}{4}\left[-\frac{\cos 4x}{4}+\frac{\cos 6x}{6}-\frac{\cos 2x}{2}\right]+\mathrm{C}.\] Final answer: $\displaystyle -\dfrac{\cos 4x}{16}+\dfrac{\cos 6x}{24}-\dfrac{\cos 2x}{8}+\mathrm{C}$.
Exercise 7
$\displaystyle \sin 4 x \sin 8 x$
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NCERT’s answer
$\displaystyle \frac{1}{2}\left[\frac{1}{4} \sin 4 x-\frac{1}{12} \sin 12 x\right]+C$
Convert the product to a sum by $\displaystyle \sin A\sin B=\tfrac12\big[\cos(A-B)-\cos(A+B)\big]$, with $\displaystyle A=4x,\;B=8x$: \[\sin 4x\sin 8x=\tfrac12\big[\cos(-4x)-\cos 12x\big]=\tfrac12\big[\cos 4x-\cos 12x\big],\] using $\displaystyle \cos(-\theta)=\cos\theta$. Integrating, \[\int\sin 4x\sin 8x\,dx=\frac{1}{2}\left[\frac{\sin 4x}{4}-\frac{\sin 12x}{12}\right]+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{\sin 4x}{8}-\dfrac{\sin 12x}{24}+\mathrm{C}$.
Exercise 8
$\displaystyle \frac{1-\cos x}{1+\cos x}$
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NCERT’s answer
$\displaystyle 2 \tan \frac{x}{2}-x+C$
Write numerator and denominator in half-angle form, using $\displaystyle 1-\cos x=2\sin^{2}\dfrac{x}{2}$ and $\displaystyle 1+\cos x=2\cos^{2}\dfrac{x}{2}$: \[\frac{1-\cos x}{1+\cos x}=\frac{2\sin^{2}(x/2)}{2\cos^{2}(x/2)}=\tan^{2}\frac{x}{2}.\] A squared tangent is never integrated directly; convert it with $\displaystyle \tan^{2}\theta=\sec^{2}\theta-1$: \[\int\tan^{2}\frac{x}{2}\,dx=\int\left(\sec^{2}\frac{x}{2}-1\right)dx=\frac{\tan(x/2)}{1/2}-x+\mathrm{C}.\] Final answer: $\displaystyle 2\tan\dfrac{x}{2}-x+\mathrm{C}$, valid wherever $\displaystyle \cos x\neq-1$.
Exercise 9
$\displaystyle \frac{\cos x}{1+\cos x}$
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NCERT’s answer
$\displaystyle x-\tan \frac{x}{2}+C$
Split the numerator so the fraction becomes a constant minus a standard piece: \[\frac{\cos x}{1+\cos x}=\frac{(1+\cos x)-1}{1+\cos x}=1-\frac{1}{1+\cos x}.\] Now use the half-angle identity $\displaystyle 1+\cos x=2\cos^{2}\dfrac{x}{2}$: \[1-\frac{1}{2\cos^{2}(x/2)}=1-\frac{1}{2}\sec^{2}\frac{x}{2}.\] Integrating, and remembering $\displaystyle \int\sec^{2}\dfrac{x}{2}\,dx=2\tan\dfrac{x}{2}$: \[\int\frac{\cos x}{1+\cos x}\,dx=x-\frac{1}{2}\cdot 2\tan\frac{x}{2}+\mathrm{C}.\] Final answer: $\displaystyle x-\tan\dfrac{x}{2}+\mathrm{C}$, valid wherever $\displaystyle \cos x\neq-1$.
Exercise 10
$\displaystyle \sin ^{4} x$
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NCERT’s answer
$\displaystyle \frac{3 x}{8}-\frac{1}{4} \sin 2 x+\frac{1}{32} \sin 4 x+C$
Reduce the fourth power in two steps with $\displaystyle \sin^{2}\theta=\dfrac{1-\cos 2\theta}{2}$. First, \[\sin^{4}x=\left(\frac{1-\cos 2x}{2}\right)^{2}=\frac{1-2\cos 2x+\cos^{2}2x}{4}.\] The step students skip: $\displaystyle \cos^{2}2x$ is still a square, so reduce it too, by $\displaystyle \cos^{2}2x=\dfrac{1+\cos 4x}{2}$: \[\sin^{4}x=\frac{1}{4}\left[1-2\cos 2x+\frac{1+\cos 4x}{2}\right]=\frac{3}{8}-\frac{\cos 2x}{2}+\frac{\cos 4x}{8}.\] Integrating, \[\int\sin^{4}x\,dx=\frac{3x}{8}-\frac{1}{2}\cdot\frac{\sin 2x}{2}+\frac{1}{8}\cdot\frac{\sin 4x}{4}+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{3x}{8}-\dfrac{\sin 2x}{4}+\dfrac{\sin 4x}{32}+\mathrm{C}$.

EXERCISE 7.3 11–20 (part 8 of 27)

Find the integrals of the functions in Exercises $\displaystyle 1$ to $\displaystyle 22$:
Exercise 11
$\displaystyle \cos ^{4} 2 x$
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NCERT’s answer
$\displaystyle \frac{3 x}{8}+\frac{1}{8} \sin 4 x+\frac{1}{64} \sin 8 x+C$
Use $\displaystyle \cos^{2}\theta=\dfrac{1+\cos 2\theta}{2}$ twice. With $\displaystyle \theta=2x$, \[\cos^{4}2x=\left(\frac{1+\cos 4x}{2}\right)^{2}=\frac{1+2\cos 4x+\cos^{2}4x}{4}.\] Reduce the leftover square, $\displaystyle \cos^{2}4x=\dfrac{1+\cos 8x}{2}$: \[\cos^{4}2x=\frac{1}{4}\left[1+2\cos 4x+\frac{1+\cos 8x}{2}\right]=\frac{3}{8}+\frac{\cos 4x}{2}+\frac{\cos 8x}{8}.\] Integrating, \[\int\cos^{4}2x\,dx=\frac{3x}{8}+\frac{1}{2}\cdot\frac{\sin 4x}{4}+\frac{1}{8}\cdot\frac{\sin 8x}{8}+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{3x}{8}+\dfrac{\sin 4x}{8}+\dfrac{\sin 8x}{64}+\mathrm{C}$.
Exercise 12
$\displaystyle \frac{\sin ^{2} x}{1+\cos x}$
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NCERT’s answer
$\displaystyle x-\sin x+\mathrm{C}$
Use the Pythagorean identity $\displaystyle \sin^{2}x=1-\cos^{2}x$ and factor the difference of squares, so the denominator cancels: \[\frac{\sin^{2}x}{1+\cos x}=\frac{1-\cos^{2}x}{1+\cos x}=\frac{(1-\cos x)(1+\cos x)}{1+\cos x}=1-\cos x,\] the cancellation being legitimate wherever $\displaystyle 1+\cos x\neq0$. Hence \[\int\frac{\sin^{2}x}{1+\cos x}\,dx=\int(1-\cos x)\,dx=x-\sin x+\mathrm{C}.\] Final answer: $\displaystyle x-\sin x+\mathrm{C}$.
Exercise 13
$\displaystyle \frac{\cos 2 x-\cos 2 \alpha}{\cos x-\cos \alpha}$
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NCERT’s answer
$\displaystyle 2(\sin x+x \cos \alpha)+\mathrm{C}$
Write both cosines of the double angle in the same form, $\displaystyle \cos 2\theta=2\cos^{2}\theta-1$: \[\cos 2x-\cos 2\alpha=(2\cos^{2}x-1)-(2\cos^{2}\alpha-1)=2(\cos^{2}x-\cos^{2}\alpha).\] Factor the difference of squares so the denominator cancels: \[\frac{2(\cos x-\cos\alpha)(\cos x+\cos\alpha)}{\cos x-\cos\alpha}=2(\cos x+\cos\alpha),\qquad \cos x\neq\cos\alpha.\] Here $\displaystyle \alpha$ is a constant, so $\displaystyle \cos\alpha$ integrates like a number: \[\int\frac{\cos 2x-\cos 2\alpha}{\cos x-\cos\alpha}\,dx=2\int(\cos x+\cos\alpha)\,dx=2\sin x+2x\cos\alpha+\mathrm{C}.\] Final answer: $\displaystyle 2(\sin x+x\cos\alpha)+\mathrm{C}$.
Exercise 14
$\displaystyle \frac{\cos x-\sin x}{1+\sin 2 x}$
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NCERT’s answer
$\displaystyle -\frac{1}{\cos x+\sin x}+C$
Recognise the denominator as a perfect square: since $\displaystyle \sin^{2}x+\cos^{2}x=1$ and $\displaystyle \sin 2x=2\sin x\cos x$, \[1+\sin 2x=\sin^{2}x+\cos^{2}x+2\sin x\cos x=(\sin x+\cos x)^{2}.\] The numerator is exactly the derivative of $\displaystyle \sin x+\cos x$, so substitute $\displaystyle t=\sin x+\cos x$, $\displaystyle dt=(\cos x-\sin x)\,dx$: \[\int\frac{\cos x-\sin x}{1+\sin 2x}\,dx=\int\frac{dt}{t^{2}}=-\frac{1}{t}+\mathrm{C}.\] Final answer: $\displaystyle -\dfrac{1}{\sin x+\cos x}+\mathrm{C}$.
Exercise 15
$\displaystyle \tan ^{3} 2 x \sec 2 x$
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NCERT’s answer
$\displaystyle \frac{1}{6} \sec ^{3} 2 x-\frac{1}{2} \sec 2 x+C$
Keep one factor of $\displaystyle \sec 2x\tan 2x$ aside, because that is the derivative pattern of $\displaystyle \sec 2x$, and convert the rest with $\displaystyle \tan^{2}\theta=\sec^{2}\theta-1$: \[\tan^{3}2x\sec 2x=\tan^{2}2x\,(\sec 2x\tan 2x)=(\sec^{2}2x-1)\,(\sec 2x\tan 2x).\] Put $\displaystyle u=\sec 2x$; then $\displaystyle du=2\sec 2x\tan 2x\,dx$, so $\displaystyle \sec 2x\tan 2x\,dx=\tfrac12\,du$: \[\int\tan^{3}2x\sec 2x\,dx=\frac{1}{2}\int(u^{2}-1)\,du=\frac{1}{2}\left[\frac{u^{3}}{3}-u\right]+\mathrm{C}.\] Substituting back $\displaystyle u=\sec 2x$: Final answer: $\displaystyle \dfrac{\sec^{3}2x}{6}-\dfrac{\sec 2x}{2}+\mathrm{C}$.
Exercise 16
$\displaystyle \tan ^{4} x$
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NCERT’s answer
$\displaystyle \frac{1}{3} \tan ^{3} x-\tan x+x+\mathrm{C}$
Peel off $\displaystyle \tan^{2}x$ using $\displaystyle \tan^{2}x=\sec^{2}x-1$, which manufactures the $\displaystyle \sec^{2}x\,dx$ needed for a substitution: \[\tan^{4}x=\tan^{2}x(\sec^{2}x-1)=\tan^{2}x\sec^{2}x-\tan^{2}x.\] For the first piece put $\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx$, giving $\displaystyle \int t^{2}dt=\dfrac{\tan^{3}x}{3}$. For the second piece apply the identity once more: \[\int\tan^{2}x\,dx=\int(\sec^{2}x-1)\,dx=\tan x-x.\] Therefore \[\int\tan^{4}x\,dx=\frac{\tan^{3}x}{3}-(\tan x-x)+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{\tan^{3}x}{3}-\tan x+x+\mathrm{C}$.
Exercise 17
$\displaystyle \frac{\sin ^{3} x+\cos ^{3} x}{\sin ^{2} x \cos ^{2} x}$
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NCERT’s answer
$\displaystyle \sec x-\operatorname{cosec} x+\mathrm{C}$
Split the fraction over the two terms of the numerator and cancel: \[\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin^{3}x}{\sin^{2}x\cos^{2}x}+\frac{\cos^{3}x}{\sin^{2}x\cos^{2}x}=\frac{\sin x}{\cos^{2}x}+\frac{\cos x}{\sin^{2}x}.\] Each piece is a standard derivative in disguise: \[\frac{\sin x}{\cos^{2}x}=\frac{1}{\cos x}\cdot\frac{\sin x}{\cos x}=\sec x\tan x,\qquad \frac{\cos x}{\sin^{2}x}=\mathrm{cosec}\,x\cot x.\] Using $\displaystyle \int\sec x\tan x\,dx=\sec x$ and $\displaystyle \int\mathrm{cosec}\,x\cot x\,dx=-\,\mathrm{cosec}\,x$ (mind that minus sign): \[\int\frac{\sin^{3}x+\cos^{3}x}{\sin^{2}x\cos^{2}x}\,dx=\sec x-\mathrm{cosec}\,x+\mathrm{C}.\] Final answer: $\displaystyle \sec x-\mathrm{cosec}\,x+\mathrm{C}$.
Exercise 18
$\displaystyle \frac{\cos 2 x+2 \sin ^{2} x}{\cos ^{2} x}$
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NCERT’s answer
$\displaystyle \tan x+\mathrm{C}$
Choose the form of $\displaystyle \cos 2x$ that matches the other term in the numerator, namely $\displaystyle \cos 2x=1-2\sin^{2}x$: \[\cos 2x+2\sin^{2}x=(1-2\sin^{2}x)+2\sin^{2}x=1.\] The whole numerator collapses to $\displaystyle 1$, so \[\int\frac{\cos 2x+2\sin^{2}x}{\cos^{2}x}\,dx=\int\frac{dx}{\cos^{2}x}=\int\sec^{2}x\,dx.\] Final answer: $\displaystyle \tan x+\mathrm{C}$.
Exercise 19
$\displaystyle \frac{1}{\sin x \cos ^{3} x}$
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NCERT’s answer
$\displaystyle \log |\tan x|+\frac{1}{2} \tan ^{2} x+C$
Force the integrand into a function of $\displaystyle \tan x$ by dividing numerator and denominator by $\displaystyle \cos^{4}x$. Since $\displaystyle \sin x\cos^{3}x=\cos^{4}x\tan x$, \[\frac{1}{\sin x\cos^{3}x}=\frac{\sec^{4}x}{\tan x}=\frac{(1+\tan^{2}x)\sec^{2}x}{\tan x},\] using $\displaystyle \sec^{2}x=1+\tan^{2}x$ on one of the two factors of $\displaystyle \sec^{2}x$ and keeping the other for the substitution. Put $\displaystyle t=\tan x,\;dt=\sec^{2}x\,dx$: \[\int\frac{dx}{\sin x\cos^{3}x}=\int\frac{1+t^{2}}{t}\,dt=\int\left(\frac{1}{t}+t\right)dt=\log|t|+\frac{t^{2}}{2}+\mathrm{C}.\] Final answer: $\displaystyle \log|\tan x|+\dfrac{\tan^{2}x}{2}+\mathrm{C}$.
Exercise 20
$\displaystyle \frac{\cos 2 x}{(\cos x+\sin x)^{2}}$
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NCERT’s answer
$\displaystyle \log |\cos x+\sin x|+\mathrm{C}$
Expand $\displaystyle \cos 2x$ as a difference of squares, $\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)$. Then one factor of the denominator cancels: \[\frac{\cos 2x}{(\cos x+\sin x)^{2}}=\frac{(\cos x-\sin x)(\cos x+\sin x)}{(\cos x+\sin x)^{2}}=\frac{\cos x-\sin x}{\cos x+\sin x}.\] The numerator is now the derivative of the denominator, so put $\displaystyle t=\cos x+\sin x$, $\displaystyle dt=(\cos x-\sin x)\,dx$: \[\int\frac{\cos 2x}{(\cos x+\sin x)^{2}}\,dx=\int\frac{dt}{t}=\log|t|+\mathrm{C}.\] Final answer: $\displaystyle \log|\cos x+\sin x|+\mathrm{C}$.

EXERCISE 7.3 21–24 (part 9 of 27)

Find the integrals of the functions in Exercises $\displaystyle 1$ to $\displaystyle 22$:
Exercise 21
$\displaystyle \sin ^{-1}(\cos x)$
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NCERT’s answer
$\displaystyle \frac{\pi x}{2}-\frac{x^{2}}{2}+\mathrm{C}$
First simplify $\displaystyle \sin^{-1}(\cos x)$; this is where the domain matters. Write $\displaystyle \cos x=\sin\left(\dfrac{\pi}{2}-x\right)$. The identity $\displaystyle \sin^{-1}(\sin\theta)=\theta$ is valid only when $\displaystyle \theta$ lies in the principal range $\displaystyle \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, and $\displaystyle \dfrac{\pi}{2}-x$ lies there exactly when $\displaystyle 0\le x\le\pi$. So, for $\displaystyle x\in[0,\pi]$, \[\sin^{-1}(\cos x)=\sin^{-1}\!\left(\sin\left(\frac{\pi}{2}-x\right)\right)=\frac{\pi}{2}-x.\] Integrating this linear function, \[\int\sin^{-1}(\cos x)\,dx=\int\left(\frac{\pi}{2}-x\right)dx=\frac{\pi x}{2}-\frac{x^{2}}{2}+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{\pi x}{2}-\dfrac{x^{2}}{2}+\mathrm{C}$, on $\displaystyle 0\le x\le\pi$ (outside that interval the simplification, and hence the formula, changes).
Exercise 22
$\displaystyle \frac{1}{\cos (x-a) \cos (x-b)}$
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NCERT’s answer
$\displaystyle \frac{1}{\sin (a-b)} \log \left|\frac{\cos (x-a)}{\cos (x-b)}\right|+\mathrm{C}$
The two angles differ by a constant: $\displaystyle (x-b)-(x-a)=a-b$. Multiply and divide by $\displaystyle \sin(a-b)$ (a non-zero constant when $\displaystyle a\neq b$) and expand it as the sine of that difference: \[\sin(a-b)=\sin\big[(x-b)-(x-a)\big]=\sin(x-b)\cos(x-a)-\cos(x-b)\sin(x-a).\] Dividing this equation by $\displaystyle \cos(x-a)\cos(x-b)$ gives \[\frac{\sin(a-b)}{\cos(x-a)\cos(x-b)}=\tan(x-b)-\tan(x-a),\] so \[\frac{1}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\big[\tan(x-b)-\tan(x-a)\big].\] Now integrate, using $\displaystyle \int\tan(x-k)\,dx=-\log|\cos(x-k)|$: \[\int\frac{dx}{\cos(x-a)\cos(x-b)}=\frac{1}{\sin(a-b)}\Big[-\log|\cos(x-b)|+\log|\cos(x-a)|\Big]+\mathrm{C}.\] Final answer: $\displaystyle \dfrac{1}{\sin(a-b)}\log\left|\dfrac{\cos(x-a)}{\cos(x-b)}\right|+\mathrm{C}$.
Choose the correct answer in Exercises $\displaystyle 23$ and 24.
Exercise 23
$\displaystyle \int \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{2} x \cos ^{2} x} d x$ is equal to (A) $\displaystyle \tan x+\cot x+\mathrm{C}$ (B) $\displaystyle \tan x+\operatorname{cosec} x+\mathrm{C}$ (C) $\displaystyle -\tan x+\cot x+\mathrm{C}$ (D) $\displaystyle \tan x+\sec x+\mathrm{C}$
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NCERT’s answer
A
Split the single fraction into two and cancel in each: \[\frac{\sin^{2}x-\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{\sin^{2}x}{\sin^{2}x\cos^{2}x}-\frac{\cos^{2}x}{\sin^{2}x\cos^{2}x}=\frac{1}{\cos^{2}x}-\frac{1}{\sin^{2}x}=\sec^{2}x-\mathrm{cosec}^{2}x.\] Use $\displaystyle \int\sec^{2}x\,dx=\tan x$ and $\displaystyle \int\mathrm{cosec}^{2}x\,dx=-\cot x$; the second minus sign meets that one and gives a plus: \[\int\left(\sec^{2}x-\mathrm{cosec}^{2}x\right)dx=\tan x-(-\cot x)+\mathrm{C}=\tan x+\cot x+\mathrm{C}.\] Final answer: option (A), $\displaystyle \tan x+\cot x+\mathrm{C}$.
Exercise 24
$\displaystyle \int \frac{e^{x}(1+x)}{\cos ^{2}\left(e^{x} x\right)} d x$ equals (A) $\displaystyle -\cot \left(e x^{x}\right)+\mathrm{C}$ (B) $\displaystyle \tan \left(x e^{x}\right)+\mathrm{C}$ (C) $\displaystyle \tan \left(e^{x}\right)+\mathrm{C}$ (D) $\displaystyle \cot \left(e^{x}\right)+\mathrm{C}$
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NCERT’s answer
B
Read the numerator as a derivative. By the product rule, \[\frac{d}{dx}\left(x e^{x}\right)=e^{x}+x e^{x}=e^{x}(1+x),\] which is exactly the numerator, and the angle in the denominator is that same $\displaystyle x e^{x}$. Substitute $\displaystyle t=x e^{x}$, so $\displaystyle dt=e^{x}(1+x)\,dx$: \[\int\frac{e^{x}(1+x)}{\cos^{2}\left(x e^{x}\right)}\,dx=\int\frac{dt}{\cos^{2}t}=\int\sec^{2}t\,dt=\tan t+\mathrm{C}.\] Putting $\displaystyle t=x e^{x}$ back: Final answer: option (B), $\displaystyle \tan\left(x e^{x}\right)+\mathrm{C}$.

EXERCISE 7.4 1–10 (part 10 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 1
$\displaystyle \frac{3 x^{2}}{x^{6}+1}$
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NCERT’s answer
$\displaystyle \tan ^{-1} x^{3}+\mathrm{C}$
The integrand becomes a standard form under a substitution, using $\displaystyle \int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}$.Put $\displaystyle t=x^{3}$. Then $\displaystyle dt=3x^{2}\,dx$, which is exactly the numerator, and $\displaystyle x^{6}=\left(x^{3}\right)^{2}=t^{2}$. \[\int \frac{3x^{2}}{x^{6}+1}\,dx=\int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}\] Restoring $\displaystyle t=x^{3}$, \[\int \frac{3x^{2}}{x^{6}+1}\,dx=\tan^{-1}\left(x^{3}\right)+\mathrm{C}\]
Exercise 2
$\displaystyle \frac{1}{\sqrt{1+4 x^{2}}}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left|2 x+\sqrt{1+4 x^{2}}\right|+\mathrm{C}$
Standard form: $\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}$.Write $\displaystyle 1+4x^{2}=1+(2x)^{2}$ and put $\displaystyle t=2x$, so $\displaystyle dt=2\,dx$, i.e. $\displaystyle dx=\frac{dt}{2}$. \[\int \frac{dx}{\sqrt{1+4x^{2}}}=\frac{1}{2}\int \frac{dt}{\sqrt{t^{2}+1^{2}}}=\frac{1}{2}\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{1+4x^{2}}}=\frac{1}{2}\log\left|2x+\sqrt{1+4x^{2}}\right|+\mathrm{C}\]
Exercise 3
$\displaystyle \frac{1}{\sqrt{(2-x)^{2}+1}}$
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NCERT’s answer
$\displaystyle \log \left|\frac{1}{2-x+\sqrt{x^{2}-4 x+5}}\right|+\mathrm{C}$
Standard form: $\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}$.Put $\displaystyle t=2-x$. Then $\displaystyle dt=-\,dx$, so $\displaystyle dx=-\,dt$ — this minus sign is the step most often dropped here. \[\int \frac{dx}{\sqrt{(2-x)^{2}+1}}=-\int \frac{dt}{\sqrt{t^{2}+1^{2}}}=-\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Since $\displaystyle (2-x)^{2}+1=x^{2}-4x+5$, \[\int \frac{dx}{\sqrt{(2-x)^{2}+1}}=-\log\left|(2-x)+\sqrt{x^{2}-4x+5}\right|+\mathrm{C}\]
Exercise 4
$\displaystyle \frac{1}{\sqrt{9-25 x^{2}}}$
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NCERT’s answer
$\displaystyle \frac{1}{5} \sin ^{-1} \frac{5 x}{3}+C$
Standard form: $\displaystyle \int \frac{dx}{\sqrt{a^{2}-x^{2}}}=\sin^{-1}\frac{x}{a}+\mathrm{C}$.The coefficient of $\displaystyle x^{2}$ must first be made $\displaystyle 1$ inside the radical: \[9-25x^{2}=25\left(\frac{9}{25}-x^{2}\right)\quad\Rightarrow\quad \sqrt{9-25x^{2}}=5\sqrt{\left(\frac{3}{5}\right)^{2}-x^{2}}\] Therefore \[\int \frac{dx}{\sqrt{9-25x^{2}}}=\frac{1}{5}\int \frac{dx}{\sqrt{\left(\dfrac{3}{5}\right)^{2}-x^{2}}}=\frac{1}{5}\sin^{-1}\!\left(\frac{x}{3/5}\right)+\mathrm{C}\] \[=\frac{1}{5}\sin^{-1}\!\left(\frac{5x}{3}\right)+\mathrm{C},\qquad |x|<\frac{3}{5}\]
Exercise 5
$\displaystyle \frac{3 x}{1+2 x^{4}}$
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NCERT’s answer
$\displaystyle \frac{3}{2 \sqrt{2}} \tan ^{-1} \sqrt{2} x^{2}+\mathrm{C}$
Substitute to turn the quartic into a quadratic, then use $\displaystyle \int \frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a}+\mathrm{C}$.Put $\displaystyle t=x^{2}$. Then $\displaystyle dt=2x\,dx$, so $\displaystyle x\,dx=\frac{dt}{2}$, and $\displaystyle x^{4}=t^{2}$. \[\int \frac{3x}{1+2x^{4}}\,dx=\frac{3}{2}\int \frac{dt}{1+2t^{2}}=\frac{3}{4}\int \frac{dt}{t^{2}+\left(\dfrac{1}{\sqrt{2}}\right)^{2}}\] (the factor $\displaystyle 2$ in the denominator is taken outside, which is where the $\displaystyle \frac{3}{4}$ comes from). \[=\frac{3}{4}\cdot\frac{1}{1/\sqrt{2}}\tan^{-1}\!\left(\frac{t}{1/\sqrt{2}}\right)+\mathrm{C}=\frac{3\sqrt{2}}{4}\tan^{-1}\left(\sqrt{2}\,t\right)+\mathrm{C}\] Hence \[\int \frac{3x}{1+2x^{4}}\,dx=\frac{3}{2\sqrt{2}}\tan^{-1}\left(\sqrt{2}\,x^{2}\right)+\mathrm{C}\]
Exercise 6
$\displaystyle \frac{x^{2}}{1-x^{6}}$
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NCERT’s answer
$\displaystyle \frac{1}{6} \log \left|\frac{1+x^{3}}{1-x^{3}}\right|+\mathrm{C}$
Standard form: $\displaystyle \int \frac{dt}{1-t^{2}}=\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|+\mathrm{C}$.Put $\displaystyle t=x^{3}$. Then $\displaystyle dt=3x^{2}\,dx$, so $\displaystyle x^{2}\,dx=\frac{dt}{3}$, and $\displaystyle x^{6}=t^{2}$. \[\int \frac{x^{2}}{1-x^{6}}\,dx=\frac{1}{3}\int \frac{dt}{1-t^{2}}=\frac{1}{3}\cdot\frac{1}{2}\log\left|\frac{1+t}{1-t}\right|+\mathrm{C}\] Hence \[\int \frac{x^{2}}{1-x^{6}}\,dx=\frac{1}{6}\log\left|\frac{1+x^{3}}{1-x^{3}}\right|+\mathrm{C}\]
Exercise 7
$\displaystyle \frac{x-1}{\sqrt{x^{2}-1}}$
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NCERT’s answer
$\displaystyle \sqrt{x^{2}-1}-\log \left|x+\sqrt{x^{2}-1}\right|+C$
Split the numerator so that one piece is a multiple of the derivative of $\displaystyle x^{2}-1$ and the other is a standard form. \[\int \frac{x-1}{\sqrt{x^{2}-1}}\,dx=\int \frac{x}{\sqrt{x^{2}-1}}\,dx-\int \frac{dx}{\sqrt{x^{2}-1}}\] For the first integral put $\displaystyle u=x^{2}-1$, so $\displaystyle du=2x\,dx$: \[\int \frac{x\,dx}{\sqrt{x^{2}-1}}=\frac{1}{2}\int u^{-1/2}\,du=\sqrt{u}=\sqrt{x^{2}-1}\] The second is the standard form $\displaystyle \int \frac{dx}{\sqrt{x^{2}-a^{2}}}=\log\left|x+\sqrt{x^{2}-a^{2}}\right|$ with $\displaystyle a=1$.Therefore \[\int \frac{x-1}{\sqrt{x^{2}-1}}\,dx=\sqrt{x^{2}-1}-\log\left|x+\sqrt{x^{2}-1}\right|+\mathrm{C}\]
Exercise 8
$\displaystyle \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}$
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NCERT’s answer
$\displaystyle \frac{1}{3} \log \left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+\mathrm{C}$
Standard form: $\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}$.Put $\displaystyle t=x^{3}$, so $\displaystyle dt=3x^{2}\,dx$, i.e. $\displaystyle x^{2}\,dx=\frac{dt}{3}$. Also $\displaystyle x^{6}=t^{2}$ and $\displaystyle a^{6}=\left(a^{3}\right)^{2}$. \[\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\,dx=\frac{1}{3}\int \frac{dt}{\sqrt{t^{2}+\left(a^{3}\right)^{2}}}=\frac{1}{3}\log\left|t+\sqrt{t^{2}+a^{6}}\right|+\mathrm{C}\] Hence \[\int \frac{x^{2}}{\sqrt{x^{6}+a^{6}}}\,dx=\frac{1}{3}\log\left|x^{3}+\sqrt{x^{6}+a^{6}}\right|+\mathrm{C}\]
Exercise 9
$\displaystyle \frac{\sec ^{2} x}{\sqrt{\tan ^{2} x+4}}$
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NCERT’s answer
$\displaystyle \log \left|\tan x+\sqrt{\tan ^{2} x+4}\right|+C$
The numerator $\displaystyle \sec^{2}x$ is the derivative of $\displaystyle \tan x$, so substitute for $\displaystyle \tan x$.Put $\displaystyle t=\tan x$. Then $\displaystyle dt=\sec^{2}x\,dx$. \[\int \frac{\sec^{2}x}{\sqrt{\tan^{2}x+4}}\,dx=\int \frac{dt}{\sqrt{t^{2}+2^{2}}}\] By $\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}$ with $\displaystyle a=2$, \[=\log\left|t+\sqrt{t^{2}+4}\right|+\mathrm{C}=\log\left|\tan x+\sqrt{\tan^{2}x+4}\right|+\mathrm{C}\]
Exercise 10
$\displaystyle \frac{1}{\sqrt{x^{2}+2 x+2}}$
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NCERT’s answer
$\displaystyle \log \left|x+1+\sqrt{x^{2}+2 x+2}\right|+\mathrm{C}$
Complete the square in the radicand, then apply $\displaystyle \int \frac{dt}{\sqrt{t^{2}+a^{2}}}=\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C}$. \[x^{2}+2x+2=\left(x^{2}+2x+1\right)+1=(x+1)^{2}+1^{2}\] Put $\displaystyle t=x+1$, so $\displaystyle dt=dx$. \[\int \frac{dx}{\sqrt{x^{2}+2x+2}}=\int \frac{dt}{\sqrt{t^{2}+1}}=\log\left|t+\sqrt{t^{2}+1}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{x^{2}+2x+2}}=\log\left|(x+1)+\sqrt{x^{2}+2x+2}\right|+\mathrm{C}\]

EXERCISE 7.4 11–20 (part 11 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 11
$\displaystyle \frac{1}{9 x^{2}+6 x+5}$
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NCERT’s answer
$\displaystyle \frac{1}{6} \tan ^{-1}\left(\frac{3 x+1}{2}\right)+\mathrm{C}$
Complete the square, then use $\displaystyle \int \frac{dt}{t^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{t}{a}+\mathrm{C}$. \[9x^{2}+6x+5=\left(9x^{2}+6x+1\right)+4=(3x+1)^{2}+2^{2}\] Put $\displaystyle t=3x+1$; then $\displaystyle dt=3\,dx$, i.e. $\displaystyle dx=\frac{dt}{3}$ — forgetting this $\displaystyle \frac{1}{3}$ is the usual slip. \[\int \frac{dx}{9x^{2}+6x+5}=\frac{1}{3}\int \frac{dt}{t^{2}+2^{2}}=\frac{1}{3}\cdot\frac{1}{2}\tan^{-1}\frac{t}{2}+\mathrm{C}\] Hence \[\int \frac{dx}{9x^{2}+6x+5}=\frac{1}{6}\tan^{-1}\!\left(\frac{3x+1}{2}\right)+\mathrm{C}\]
Exercise 12
$\displaystyle \frac{1}{\sqrt{7-6 x-x^{2}}}$
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NCERT’s answer
$\displaystyle \sin ^{-1}\left(\frac{x+3}{4}\right)+\mathrm{C}$
Complete the square. Because the coefficient of $\displaystyle x^{2}$ is negative, take $\displaystyle -1$ out first. \[7-6x-x^{2}=7-\left(x^{2}+6x\right)=7-\left[(x+3)^{2}-9\right]=16-(x+3)^{2}=4^{2}-(x+3)^{2}\] Put $\displaystyle t=x+3$, $\displaystyle dt=dx$, and use $\displaystyle \int \frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}$. \[\int \frac{dx}{\sqrt{7-6x-x^{2}}}=\int \frac{dt}{\sqrt{4^{2}-t^{2}}}=\sin^{-1}\frac{t}{4}+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{7-6x-x^{2}}}=\sin^{-1}\!\left(\frac{x+3}{4}\right)+\mathrm{C},\qquad -7<x<1\]
Exercise 13
$\displaystyle \frac{1}{\sqrt{(x-1)(x-2)}}$
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NCERT’s answer
$\displaystyle \log \left|x-\frac{3}{2}+\sqrt{x^{2}-3 x+2}\right|+\mathrm{C}$
Expand the product first, then complete the square — the factored form is not itself a standard form. \[(x-1)(x-2)=x^{2}-3x+2=\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}+2=\left(x-\frac{3}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\] Put $\displaystyle t=x-\frac{3}{2}$, $\displaystyle dt=dx$, and use $\displaystyle \int \frac{dt}{\sqrt{t^{2}-a^{2}}}=\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}$. \[\int \frac{dx}{\sqrt{(x-1)(x-2)}}=\int \frac{dt}{\sqrt{t^{2}-\left(\dfrac{1}{2}\right)^{2}}}=\log\left|t+\sqrt{t^{2}-\tfrac{1}{4}}\right|+\mathrm{C}\] Hence \[\int \frac{dx}{\sqrt{(x-1)(x-2)}}=\log\left|\left(x-\frac{3}{2}\right)+\sqrt{x^{2}-3x+2}\right|+\mathrm{C}\] valid where the radicand is positive, i.e. for $\displaystyle x>2$ or $\displaystyle x<1$.
Exercise 14
$\displaystyle \frac{1}{\sqrt{8+3 x-x^{2}}}$
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NCERT’s answer
$\displaystyle \sin ^{-1}\left(\frac{2 x-3}{\sqrt{41}}\right)+C$
Complete the square, taking $\displaystyle -1$ out of the $\displaystyle x$-terms because the $\displaystyle x^{2}$ coefficient is negative. \[8+3x-x^{2}=8-\left(x^{2}-3x\right)=8-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]=\frac{41}{4}-\left(x-\frac{3}{2}\right)^{2}\] So the radicand is $\displaystyle \left(\frac{\sqrt{41}}{2}\right)^{2}-\left(x-\frac{3}{2}\right)^{2}$. Put $\displaystyle t=x-\frac{3}{2}$, $\displaystyle dt=dx$, and use $\displaystyle \int \frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}$. \[\int \frac{dx}{\sqrt{8+3x-x^{2}}}=\sin^{-1}\!\left(\frac{x-\dfrac{3}{2}}{\dfrac{\sqrt{41}}{2}}\right)+\mathrm{C}=\sin^{-1}\!\left(\frac{2x-3}{\sqrt{41}}\right)+\mathrm{C}\]
Exercise 15
$\displaystyle \frac{1}{\sqrt{(x-a)(x-b)}}$
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NCERT’s answer
$\displaystyle \log \left|x-\frac{a+b}{2}+\sqrt{(x-a)(x-b)}\right|+\mathrm{C}$
Multiply out and complete the square; the half-sum $\displaystyle \frac{a+b}{2}$ is the shift and the half-difference $\displaystyle \frac{a-b}{2}$ is the constant. \[(x-a)(x-b)=x^{2}-(a+b)x+ab=\left(x-\frac{a+b}{2}\right)^{2}-\frac{(a+b)^{2}}{4}+ab\] and $\displaystyle \dfrac{(a+b)^{2}}{4}-ab=\dfrac{(a-b)^{2}}{4}$, so \[(x-a)(x-b)=\left(x-\frac{a+b}{2}\right)^{2}-\left(\frac{a-b}{2}\right)^{2}\] Put $\displaystyle t=x-\frac{a+b}{2}$, $\displaystyle dt=dx$, and use $\displaystyle \int \frac{dt}{\sqrt{t^{2}-c^{2}}}=\log\left|t+\sqrt{t^{2}-c^{2}}\right|+\mathrm{C}$. \[\int \frac{dx}{\sqrt{(x-a)(x-b)}}=\log\left|\left(x-\frac{a+b}{2}\right)+\sqrt{(x-a)(x-b)}\right|+\mathrm{C}\] valid on the intervals where $\displaystyle (x-a)(x-b)>0$.
Exercise 16
$\displaystyle \frac{4 x+1}{\sqrt{2 x^{2}+x-3}}$
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NCERT’s answer
$\displaystyle 2 \sqrt{2 x^{2}+x-3}+C$
Check the numerator against the derivative of the radicand before doing anything else. \[\frac{d}{dx}\left(2x^{2}+x-3\right)=4x+1,\] which is exactly the numerator. So put $\displaystyle u=2x^{2}+x-3$, giving $\displaystyle du=(4x+1)\,dx$. \[\int \frac{4x+1}{\sqrt{2x^{2}+x-3}}\,dx=\int \frac{du}{\sqrt{u}}=\int u^{-1/2}\,du=2\sqrt{u}+\mathrm{C}\] Hence \[\int \frac{4x+1}{\sqrt{2x^{2}+x-3}}\,dx=2\sqrt{2x^{2}+x-3}+\mathrm{C}\]
Exercise 17
$\displaystyle \frac{x+2}{\sqrt{x^{2}-1}}$
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NCERT’s answer
$\displaystyle \sqrt{x^{2}-1}+2 \log \left|x+\sqrt{x^{2}-1}\right|+C$
Split the numerator into a multiple of $\displaystyle \frac{d}{dx}\left(x^{2}-1\right)=2x$ plus a constant: \[x+2=\frac{1}{2}(2x)+2\] \[\int \frac{x+2}{\sqrt{x^{2}-1}}\,dx=\frac{1}{2}\int \frac{2x}{\sqrt{x^{2}-1}}\,dx+2\int \frac{dx}{\sqrt{x^{2}-1}}\] For the first, $\displaystyle u=x^{2}-1$, $\displaystyle du=2x\,dx$, so $\displaystyle \frac{1}{2}\int u^{-1/2}du=\sqrt{x^{2}-1}$. The second is the standard form $\displaystyle \int \frac{dx}{\sqrt{x^{2}-a^{2}}}=\log\left|x+\sqrt{x^{2}-a^{2}}\right|$ with $\displaystyle a=1$.Therefore \[\int \frac{x+2}{\sqrt{x^{2}-1}}\,dx=\sqrt{x^{2}-1}+2\log\left|x+\sqrt{x^{2}-1}\right|+\mathrm{C}\]
Exercise 18
$\displaystyle \frac{5 x-2}{1+2 x+3 x^{2}}$
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NCERT’s answer
$\displaystyle \frac{5}{6} \log \left|3 x^{2}+2 x+1\right|-\frac{11}{3 \sqrt{2}} \tan ^{-1}\left(\frac{3 x+1}{\sqrt{2}}\right)+\mathrm{C}$
Write the numerator as $\displaystyle \mathrm{A}\times$(derivative of the denominator)$\displaystyle +\mathrm{B}$.Here $\displaystyle \frac{d}{dx}\left(3x^{2}+2x+1\right)=6x+2$, so let \[5x-2=\mathrm{A}(6x+2)+\mathrm{B}\] Comparing coefficients of $\displaystyle x$: $\displaystyle 6\mathrm{A}=5\Rightarrow \mathrm{A}=\frac{5}{6}$. Comparing constants: $\displaystyle 2\mathrm{A}+\mathrm{B}=-2\Rightarrow \mathrm{B}=-2-\frac{5}{3}=-\frac{11}{3}$. \[\int \frac{5x-2}{3x^{2}+2x+1}\,dx=\frac{5}{6}\int \frac{6x+2}{3x^{2}+2x+1}\,dx-\frac{11}{3}\int \frac{dx}{3x^{2}+2x+1}\] The first integral is $\displaystyle \frac{5}{6}\log\left(3x^{2}+2x+1\right)$ (the quadratic has discriminant $\displaystyle 4-12<0$, so it is always positive and no modulus is needed).For the second, complete the square after taking out the leading $\displaystyle 3$: \[3x^{2}+2x+1=3\left[\left(x+\frac{1}{3}\right)^{2}+\frac{2}{9}\right]\] \[\int \frac{dx}{3x^{2}+2x+1}=\frac{1}{3}\int \frac{dx}{\left(x+\dfrac{1}{3}\right)^{2}+\left(\dfrac{\sqrt{2}}{3}\right)^{2}}=\frac{1}{3}\cdot\frac{3}{\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)=\frac{1}{\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)\] Therefore \[\int \frac{5x-2}{1+2x+3x^{2}}\,dx=\frac{5}{6}\log\left(3x^{2}+2x+1\right)-\frac{11}{3\sqrt{2}}\tan^{-1}\!\left(\frac{3x+1}{\sqrt{2}}\right)+\mathrm{C}\]
Exercise 19
$\displaystyle \frac{6 x+7}{\sqrt{(x-5)(x-4)}}$
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NCERT’s answer
$\displaystyle 6 \sqrt{x^{2}-9 x+20}+34 \log \left|x-\frac{9}{2}+\sqrt{x^{2}-9 x+20}\right|+\mathrm{C}$
First expand the product: $\displaystyle (x-5)(x-4)=x^{2}-9x+20$, whose derivative is $\displaystyle 2x-9$.Write $\displaystyle 6x+7=\mathrm{A}(2x-9)+\mathrm{B}$. Coefficients of $\displaystyle x$: $\displaystyle 2\mathrm{A}=6\Rightarrow \mathrm{A}=3$. Constants: $\displaystyle -9\mathrm{A}+\mathrm{B}=7\Rightarrow \mathrm{B}=7+27=34$. \[\int \frac{6x+7}{\sqrt{x^{2}-9x+20}}\,dx=3\int \frac{2x-9}{\sqrt{x^{2}-9x+20}}\,dx+34\int \frac{dx}{\sqrt{x^{2}-9x+20}}\] For the first, $\displaystyle u=x^{2}-9x+20$, $\displaystyle du=(2x-9)\,dx$, giving $\displaystyle 3\int u^{-1/2}du=6\sqrt{x^{2}-9x+20}$.For the second, complete the square: \[x^{2}-9x+20=\left(x-\frac{9}{2}\right)^{2}-\frac{81}{4}+20=\left(x-\frac{9}{2}\right)^{2}-\left(\frac{1}{2}\right)^{2}\] so $\displaystyle \int \frac{dx}{\sqrt{x^{2}-9x+20}}=\log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9x+20}\right|$.Therefore \[\int \frac{6x+7}{\sqrt{(x-5)(x-4)}}\,dx=6\sqrt{x^{2}-9x+20}+34\log\left|\left(x-\frac{9}{2}\right)+\sqrt{x^{2}-9x+20}\right|+\mathrm{C}\]
Exercise 20
$\displaystyle \frac{x+2}{\sqrt{4 x-x^{2}}}$
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NCERT’s answer
$\displaystyle -\sqrt{4 x-x^{2}}+4 \sin ^{-1}\left(\frac{x-2}{2}\right)+\mathrm{C}$
The derivative of $\displaystyle 4x-x^{2}$ is $\displaystyle 4-2x$, so split the numerator against that.Let $\displaystyle x+2=\mathrm{A}(4-2x)+\mathrm{B}$. Coefficients of $\displaystyle x$: $\displaystyle -2\mathrm{A}=1\Rightarrow \mathrm{A}=-\frac{1}{2}$. Constants: $\displaystyle 4\mathrm{A}+\mathrm{B}=2\Rightarrow \mathrm{B}=4$. \[\int \frac{x+2}{\sqrt{4x-x^{2}}}\,dx=-\frac{1}{2}\int \frac{4-2x}{\sqrt{4x-x^{2}}}\,dx+4\int \frac{dx}{\sqrt{4x-x^{2}}}\] First part: with $\displaystyle u=4x-x^{2}$, $\displaystyle du=(4-2x)\,dx$, so $\displaystyle -\frac{1}{2}\int u^{-1/2}du=-\sqrt{4x-x^{2}}$.Second part: complete the square, $\displaystyle 4x-x^{2}=4-(x-2)^{2}=2^{2}-(x-2)^{2}$, so \[4\int \frac{dx}{\sqrt{2^{2}-(x-2)^{2}}}=4\sin^{-1}\!\left(\frac{x-2}{2}\right)\] Therefore \[\int \frac{x+2}{\sqrt{4x-x^{2}}}\,dx=-\sqrt{4x-x^{2}}+4\sin^{-1}\!\left(\frac{x-2}{2}\right)+\mathrm{C},\qquad 0<x<4\]

EXERCISE 7.4 21–25 (part 12 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 21
$\displaystyle \frac{x+2}{\sqrt{x^{2}+2 x+3}}$
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NCERT’s answer
$\displaystyle \sqrt{x^{2}+2 x+3}+\log \left|x+1+\sqrt{x^{2}+2 x+3}\right|+\mathrm{C}$
Split the numerator against $\displaystyle \frac{d}{dx}\left(x^{2}+2x+3\right)=2x+2$: \[x+2=\frac{1}{2}(2x+2)+1\] \[\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx=\frac{1}{2}\int \frac{2x+2}{\sqrt{x^{2}+2x+3}}\,dx+\int \frac{dx}{\sqrt{x^{2}+2x+3}}\] First part: with $\displaystyle u=x^{2}+2x+3$, $\displaystyle du=(2x+2)\,dx$, it is $\displaystyle \frac{1}{2}\cdot 2\sqrt{u}=\sqrt{x^{2}+2x+3}$.Second part: complete the square, $\displaystyle x^{2}+2x+3=(x+1)^{2}+\left(\sqrt{2}\right)^{2}$, so \[\int \frac{dx}{\sqrt{(x+1)^{2}+2}}=\log\left|(x+1)+\sqrt{x^{2}+2x+3}\right|\] Therefore \[\int \frac{x+2}{\sqrt{x^{2}+2x+3}}\,dx=\sqrt{x^{2}+2x+3}+\log\left|(x+1)+\sqrt{x^{2}+2x+3}\right|+\mathrm{C}\]
Exercise 22
$\displaystyle \frac{x+3}{x^{2}-2 x-5}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left|x^{2}-2 x-5\right|+\frac{2}{\sqrt{6}} \log \left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+\mathrm{C}$
Split the numerator against $\displaystyle \frac{d}{dx}\left(x^{2}-2x-5\right)=2x-2$: \[x+3=\frac{1}{2}(2x-2)+4\] \[\int \frac{x+3}{x^{2}-2x-5}\,dx=\frac{1}{2}\int \frac{2x-2}{x^{2}-2x-5}\,dx+4\int \frac{dx}{x^{2}-2x-5}\] First part: $\displaystyle \frac{1}{2}\log\left|x^{2}-2x-5\right|$.Second part: complete the square. The constant is $\displaystyle -6$, i.e. negative, so this is the $\displaystyle t^{2}-a^{2}$ form, not an arctangent: \[x^{2}-2x-5=(x-1)^{2}-6=(x-1)^{2}-\left(\sqrt{6}\right)^{2}\] Using $\displaystyle \int \frac{dt}{t^{2}-a^{2}}=\frac{1}{2a}\log\left|\frac{t-a}{t+a}\right|+\mathrm{C}$, \[4\int \frac{dx}{(x-1)^{2}-6}=4\cdot\frac{1}{2\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|=\frac{2}{\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|\] Therefore \[\int \frac{x+3}{x^{2}-2x-5}\,dx=\frac{1}{2}\log\left|x^{2}-2x-5\right|+\frac{2}{\sqrt{6}}\log\left|\frac{x-1-\sqrt{6}}{x-1+\sqrt{6}}\right|+\mathrm{C}\]
Exercise 23
$\displaystyle \frac{5 x+3}{\sqrt{x^{2}+4 x+10}}$.
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NCERT’s answer
$\displaystyle 5 \sqrt{x^{2}+4 x+10}-7 \log \left|x+2+\sqrt{x^{2}+4 x+10}\right|+\mathrm{C}$
Split the numerator against $\displaystyle \frac{d}{dx}\left(x^{2}+4x+10\right)=2x+4$.Let $\displaystyle 5x+3=\mathrm{A}(2x+4)+\mathrm{B}$. Coefficients of $\displaystyle x$: $\displaystyle 2\mathrm{A}=5\Rightarrow \mathrm{A}=\frac{5}{2}$. Constants: $\displaystyle 4\mathrm{A}+\mathrm{B}=3\Rightarrow \mathrm{B}=3-10=-7$. \[\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\,dx=\frac{5}{2}\int \frac{2x+4}{\sqrt{x^{2}+4x+10}}\,dx-7\int \frac{dx}{\sqrt{x^{2}+4x+10}}\] First part: with $\displaystyle u=x^{2}+4x+10$, $\displaystyle du=(2x+4)\,dx$, it is $\displaystyle \frac{5}{2}\cdot 2\sqrt{u}=5\sqrt{x^{2}+4x+10}$.Second part: $\displaystyle x^{2}+4x+10=(x+2)^{2}+6=(x+2)^{2}+\left(\sqrt{6}\right)^{2}$, so \[\int \frac{dx}{\sqrt{(x+2)^{2}+6}}=\log\left|(x+2)+\sqrt{x^{2}+4x+10}\right|\] Therefore \[\int \frac{5x+3}{\sqrt{x^{2}+4x+10}}\,dx=5\sqrt{x^{2}+4x+10}-7\log\left|(x+2)+\sqrt{x^{2}+4x+10}\right|+\mathrm{C}\]
Choose the correct answer in Exercises $\displaystyle 24$ and 25.
Exercise 24
$\displaystyle \int \frac{d x}{x^{2}+2 x+2}$ equals (A) $\displaystyle x \tan ^{-1}(x+1)+\mathrm{C}$ (B) $\displaystyle \tan ^{-1}(x+1)+\mathrm{C}$ (C) $\displaystyle (x+1) \tan ^{-1} x+\mathrm{C}$ (D) $\displaystyle \tan ^{-1} x+\mathrm{C}$
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NCERT’s answer
B
Complete the square in the denominator: \[x^{2}+2x+2=(x+1)^{2}+1\] Put $\displaystyle t=x+1$, so $\displaystyle dt=dx$, and use $\displaystyle \int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}$. \[\int \frac{dx}{x^{2}+2x+2}=\int \frac{dt}{t^{2}+1}=\tan^{-1}t+\mathrm{C}=\tan^{-1}(x+1)+\mathrm{C}\] The correct answer is (B) $\displaystyle \tan^{-1}(x+1)+\mathrm{C}$.
Exercise 25
$\displaystyle \int \frac{d x}{\sqrt{9 x-4 x^{2}}}$ equals (A) $\displaystyle \frac{1}{9} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$ (B) $\displaystyle \frac{1}{2} \sin ^{-1}\left(\frac{8 x-9}{9}\right)+\mathrm{C}$ (C) $\displaystyle \frac{1}{3} \sin ^{-1}\left(\frac{9 x-8}{8}\right)+C$ (D) $\displaystyle \frac{1}{2} \sin ^{-1}\left(\frac{9 x-8}{9}\right)+C$
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NCERT’s answer
B
Complete the square, remembering to factor out the coefficient $\displaystyle 4$ of $\displaystyle x^{2}$ first — that factor is what produces the $\displaystyle \frac{1}{2}$ outside. \[9x-4x^{2}=-4\left(x^{2}-\frac{9}{4}x\right)=-4\left[\left(x-\frac{9}{8}\right)^{2}-\frac{81}{64}\right]=4\left[\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}\right]\] So $\displaystyle \sqrt{9x-4x^{2}}=2\sqrt{\left(\frac{9}{8}\right)^{2}-\left(x-\frac{9}{8}\right)^{2}}$, and with $\displaystyle t=x-\frac{9}{8}$, $\displaystyle dt=dx$, \[\int \frac{dx}{\sqrt{9x-4x^{2}}}=\frac{1}{2}\int \frac{dt}{\sqrt{\left(\dfrac{9}{8}\right)^{2}-t^{2}}}=\frac{1}{2}\sin^{-1}\!\left(\frac{t}{9/8}\right)+\mathrm{C}\] \[=\frac{1}{2}\sin^{-1}\!\left(\frac{x-\dfrac{9}{8}}{\dfrac{9}{8}}\right)+\mathrm{C}=\frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+\mathrm{C}\] The correct answer is (B) $\displaystyle \frac{1}{2}\sin^{-1}\!\left(\frac{8x-9}{9}\right)+\mathrm{C}$.

EXERCISE 7.5 1–10 (part 13 of 27)

Integrate the rational functions in Exercises $\displaystyle 1$ to 21.
Exercise 1
$\displaystyle \frac{x}{(x+1)(x+2)}$
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NCERT’s answer
$\displaystyle \log \frac{(x+2)^{2}}{|x+1|}+C$
The denominator has two distinct linear factors, so use partial fractions: \[\frac{x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2} \] Clearing denominators, $\displaystyle x=A(x+2)+B(x+1) $.Put $\displaystyle x=-1 $: $\displaystyle -1=A(1)\Rightarrow A=-1 $. Put $\displaystyle x=-2 $: $\displaystyle -2=B(-1)\Rightarrow B=2 $.\[\int\frac{x\,dx}{(x+1)(x+2)}=\int\left(\frac{-1}{x+1}+\frac{2}{x+2}\right)dx \] \[=-\log|x+1|+2\log|x+2|+\mathrm{C} \]
Exercise 2
$\displaystyle \frac{1}{x^{2}-9}$
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NCERT’s answer
$\displaystyle \frac{1}{6} \log \left|\frac{x-3}{x+3}\right|+\mathrm{C}$
Factorise: $\displaystyle x^{2}-9=(x-3)(x+3) $, two distinct linear factors. \[\frac{1}{x^{2}-9}=\frac{A}{x-3}+\frac{B}{x+3},\qquad 1=A(x+3)+B(x-3) \] Put $\displaystyle x=3 $: $\displaystyle 1=6A\Rightarrow A=\tfrac16 $. Put $\displaystyle x=-3 $: $\displaystyle 1=-6B\Rightarrow B=-\tfrac16 $.\[\int\frac{dx}{x^{2}-9}=\frac16\int\left(\frac{1}{x-3}-\frac{1}{x+3}\right)dx=\frac16\log\left|\frac{x-3}{x+3}\right|+\mathrm{C} \]
Exercise 3
$\displaystyle \frac{3 x-1}{(x-1)(x-2)(x-3)}$
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NCERT’s answer
$\displaystyle \log |x-1|-5 \log |x-2|+4 \log |x-3|+C$
Three distinct linear factors: \[\frac{3x-1}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \] \[3x-1=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \] Put $\displaystyle x=1 $: $\displaystyle 2=A(-1)(-2)=2A\Rightarrow A=1 $. Put $\displaystyle x=2 $: $\displaystyle 5=B(1)(-1)\Rightarrow B=-5 $. Put $\displaystyle x=3 $: $\displaystyle 8=C(2)(1)\Rightarrow C=4 $.\[\int\frac{(3x-1)\,dx}{(x-1)(x-2)(x-3)}=\log|x-1|-5\log|x-2|+4\log|x-3|+\mathrm{C} \]
Exercise 4
$\displaystyle \frac{x}{(x-1)(x-2)(x-3)}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log |x-1|-2 \log |x-2|+\frac{3}{2} \log |x-3|+\mathrm{C}$
Three distinct linear factors: \[\frac{x}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3} \] \[x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2) \] Put $\displaystyle x=1 $: $\displaystyle 1=A(-1)(-2)=2A\Rightarrow A=\tfrac12 $. Put $\displaystyle x=2 $: $\displaystyle 2=B(1)(-1)\Rightarrow B=-2 $. Put $\displaystyle x=3 $: $\displaystyle 3=C(2)(1)\Rightarrow C=\tfrac32 $.\[\int\frac{x\,dx}{(x-1)(x-2)(x-3)}=\frac12\log|x-1|-2\log|x-2|+\frac32\log|x-3|+\mathrm{C} \]
Exercise 5
$\displaystyle \frac{2 x}{x^{2}+3 x+2}$
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NCERT’s answer
$\displaystyle 4 \log |x+2|-2 \log |x+1|+C$
Factorise the denominator: $\displaystyle x^{2}+3x+2=(x+1)(x+2) $. \[\frac{2x}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2},\qquad 2x=A(x+2)+B(x+1) \] Put $\displaystyle x=-1 $: $\displaystyle -2=A(1)\Rightarrow A=-2 $. Put $\displaystyle x=-2 $: $\displaystyle -4=B(-1)\Rightarrow B=4 $.\[\int\frac{2x\,dx}{x^{2}+3x+2}=-2\log|x+1|+4\log|x+2|+\mathrm{C} \]
Exercise 6
$\displaystyle \frac{1-x^{2}}{x(1-2 x)}$
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NCERT’s answer
$\displaystyle \frac{x}{2}+\log |x|-\frac{3}{4} \log |1-2 x|+C$
The integrand is an improper rational function: the numerator $\displaystyle 1-x^{2} $ and the denominator $\displaystyle x(1-2x)=x-2x^{2} $ both have degree $\displaystyle 2$, so divide before splitting.Since $\displaystyle \tfrac12\left(-2x^{2}+x\right)=-x^{2}+\tfrac{x}{2} $, the remainder is $\displaystyle 1-\tfrac{x}{2} $: \[\frac{1-x^{2}}{x(1-2x)}=\frac12+\frac{1-\dfrac{x}{2}}{x(1-2x)}=\frac12+\frac{2-x}{2\,x(1-2x)} \]Now split the proper part: \[\frac{2-x}{x(1-2x)}=\frac{A}{x}+\frac{B}{1-2x},\qquad 2-x=A(1-2x)+Bx \] Put $\displaystyle x=0 $: $\displaystyle A=2 $. Put $\displaystyle x=\tfrac12 $: $\displaystyle \tfrac32=\tfrac{B}{2}\Rightarrow B=3 $.So the integrand is $\displaystyle \dfrac12+\dfrac1x+\dfrac{3}{2(1-2x)} $. Using $\displaystyle \displaystyle\int\frac{dx}{1-2x}=-\frac12\log|1-2x| $, \[\int\frac{(1-x^{2})\,dx}{x(1-2x)}=\frac{x}{2}+\log|x|-\frac34\log|1-2x|+\mathrm{C} \]
Exercise 7
$\displaystyle \frac{x}{\left(x^{2}+1\right)(x-1)}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log |x-1|-\frac{1}{4} \log \left(x^{2}+1\right)+\frac{1}{2} \tan ^{-1} x+C$
$\displaystyle x^{2}+1 $ is an irreducible quadratic factor, so it takes a linear numerator: \[\frac{x}{(x^{2}+1)(x-1)}=\frac{Ax+B}{x^{2}+1}+\frac{C}{x-1} \] \[x=(Ax+B)(x-1)+C(x^{2}+1) \] Put $\displaystyle x=1 $: $\displaystyle 1=2C\Rightarrow C=\tfrac12 $. Coefficient of $\displaystyle x^{2} $: $\displaystyle 0=A+C\Rightarrow A=-\tfrac12 $. Constant term: $\displaystyle 0=-B+C\Rightarrow B=\tfrac12 $.\[\int\frac{x\,dx}{(x^{2}+1)(x-1)}=-\frac12\int\frac{x\,dx}{x^{2}+1}+\frac12\int\frac{dx}{x^{2}+1}+\frac12\int\frac{dx}{x-1} \] \[=-\frac14\log\left(x^{2}+1\right)+\frac12\tan^{-1}x+\frac12\log|x-1|+\mathrm{C} \]
Exercise 8
$\displaystyle \frac{x}{(x-1)^{2}(x+2)}$
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NCERT’s answer
$\displaystyle \frac{2}{9} \log \left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+\mathrm{C}$
$\displaystyle (x-1) $ is a repeated factor, so both powers must appear: \[\frac{x}{(x-1)^{2}(x+2)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+2} \] \[x=A(x-1)(x+2)+B(x+2)+C(x-1)^{2} \] Put $\displaystyle x=1 $: $\displaystyle 1=3B\Rightarrow B=\tfrac13 $. Put $\displaystyle x=-2 $: $\displaystyle -2=9C\Rightarrow C=-\tfrac29 $. Coefficient of $\displaystyle x^{2} $: $\displaystyle 0=A+C\Rightarrow A=\tfrac29 $.\[\int\frac{x\,dx}{(x-1)^{2}(x+2)}=\frac29\log|x-1|+\frac13\cdot\left(-\frac{1}{x-1}\right)-\frac29\log|x+2|+\mathrm{C} \] \[=\frac29\log\left|\frac{x-1}{x+2}\right|-\frac{1}{3(x-1)}+\mathrm{C} \]
Exercise 9
$\displaystyle \frac{3 x+5}{x^{3}-x^{2}-x+1}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+\mathrm{C}$
Factorise the denominator by grouping: \[x^{3}-x^{2}-x+1=x^{2}(x-1)-(x-1)=(x-1)\left(x^{2}-1\right)=(x-1)^{2}(x+1) \] A repeated factor $\displaystyle (x-1) $ and a simple factor $\displaystyle (x+1) $: \[\frac{3x+5}{(x-1)^{2}(x+1)}=\frac{A}{x-1}+\frac{B}{(x-1)^{2}}+\frac{C}{x+1} \] \[3x+5=A(x-1)(x+1)+B(x+1)+C(x-1)^{2} \] Put $\displaystyle x=1 $: $\displaystyle 8=2B\Rightarrow B=4 $. Put $\displaystyle x=-1 $: $\displaystyle 2=4C\Rightarrow C=\tfrac12 $. Coefficient of $\displaystyle x^{2} $: $\displaystyle 0=A+C\Rightarrow A=-\tfrac12 $.\[\int\frac{(3x+5)\,dx}{x^{3}-x^{2}-x+1}=-\frac12\log|x-1|-\frac{4}{x-1}+\frac12\log|x+1|+\mathrm{C} \] \[=\frac12\log\left|\frac{x+1}{x-1}\right|-\frac{4}{x-1}+\mathrm{C} \]
Exercise 10
$\displaystyle \frac{2 x-3}{\left(x^{2}-1\right)(2 x+3)}$
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NCERT’s answer
$\displaystyle \frac{5}{2} \log |x+1|-\frac{1}{10} \log |x-1|-\frac{12}{5} \log |2 x+3|+C$
Since $\displaystyle x^{2}-1=(x-1)(x+1) $, there are three distinct linear factors: \[\frac{2x-3}{(x-1)(x+1)(2x+3)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{2x+3} \] \[2x-3=A(x+1)(2x+3)+B(x-1)(2x+3)+C(x-1)(x+1) \] Put $\displaystyle x=1 $: $\displaystyle -1=A(2)(5)=10A\Rightarrow A=-\tfrac{1}{10} $. Put $\displaystyle x=-1 $: $\displaystyle -5=B(-2)(1)=-2B\Rightarrow B=\tfrac52 $. Put $\displaystyle x=-\tfrac32 $ (the zero of the third factor): $\displaystyle -6=C\left(-\tfrac52\right)\left(-\tfrac12\right)=\tfrac54C\Rightarrow C=-\tfrac{24}{5} $.Note $\displaystyle \displaystyle\int\frac{dx}{2x+3}=\frac12\log|2x+3| $, so the third term contributes $\displaystyle -\tfrac{24}{5}\cdot\tfrac12=-\tfrac{12}{5} $ times the log. \[\int\frac{(2x-3)\,dx}{\left(x^{2}-1\right)(2x+3)}=-\frac{1}{10}\log|x-1|+\frac52\log|x+1|-\frac{12}{5}\log|2x+3|+\mathrm{C} \]

EXERCISE 7.5 11–23 (part 14 of 27)

Integrate the rational functions in Exercises $\displaystyle 1$ to 21.
Exercise 11
$\displaystyle \frac{5 x}{(x+1)\left(x^{2}-4\right)}$
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NCERT’s answer
$\displaystyle \frac{5}{3} \log |x+1|-\frac{5}{2} \log |x+2|+\frac{5}{6} \log |x-2|+\mathrm{C}$
Since $\displaystyle x^{2}-4=(x-2)(x+2) $, there are three distinct linear factors: \[\frac{5x}{(x+1)(x-2)(x+2)}=\frac{A}{x+1}+\frac{B}{x-2}+\frac{C}{x+2} \] \[5x=A(x-2)(x+2)+B(x+1)(x+2)+C(x+1)(x-2) \] Put $\displaystyle x=-1 $: $\displaystyle -5=A(-3)(1)=-3A\Rightarrow A=\tfrac53 $. Put $\displaystyle x=2 $: $\displaystyle 10=B(3)(4)=12B\Rightarrow B=\tfrac56 $. Put $\displaystyle x=-2 $: $\displaystyle -10=C(-1)(-4)=4C\Rightarrow C=-\tfrac52 $.\[\int\frac{5x\,dx}{(x+1)\left(x^{2}-4\right)}=\frac53\log|x+1|+\frac56\log|x-2|-\frac52\log|x+2|+\mathrm{C} \]
Exercise 12
$\displaystyle \frac{x^{3}+x+1}{x^{2}-1}$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2}+\frac{1}{2} \log |x+1|+\frac{3}{2} \log |x-1|+C$
The integrand is improper (degree $\displaystyle 3$ over degree $\displaystyle 2$), so divide first. Since $\displaystyle x^{3}+x+1=x\left(x^{2}-1\right)+2x+1 $, \[\frac{x^{3}+x+1}{x^{2}-1}=x+\frac{2x+1}{x^{2}-1} \] Now split the proper part: \[\frac{2x+1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1},\qquad 2x+1=A(x+1)+B(x-1) \] Put $\displaystyle x=1 $: $\displaystyle 3=2A\Rightarrow A=\tfrac32 $. Put $\displaystyle x=-1 $: $\displaystyle -1=-2B\Rightarrow B=\tfrac12 $.\[\int\frac{x^{3}+x+1}{x^{2}-1}\,dx=\frac{x^{2}}{2}+\frac32\log|x-1|+\frac12\log|x+1|+\mathrm{C} \]
Exercise 13
$\displaystyle \frac{2}{(1-x)\left(1+x^{2}\right)}$
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NCERT’s answer
$\displaystyle -\log |x-1|+\frac{1}{2} \log \left(1+x^{2}\right)+\tan ^{-1} x+\mathrm{C}$
$\displaystyle 1+x^{2} $ is irreducible, so it takes a linear numerator: \[\frac{2}{(1-x)\left(1+x^{2}\right)}=\frac{A}{1-x}+\frac{Bx+C}{1+x^{2}} \] \[2=A\left(1+x^{2}\right)+(Bx+C)(1-x) \] Put $\displaystyle x=1 $: $\displaystyle 2=2A\Rightarrow A=1 $. Coefficient of $\displaystyle x^{2} $: $\displaystyle 0=A-B\Rightarrow B=1 $. Constant term: $\displaystyle 2=A+C\Rightarrow C=1 $.So the integrand is $\displaystyle \dfrac{1}{1-x}+\dfrac{x}{1+x^{2}}+\dfrac{1}{1+x^{2}} $. Note $\displaystyle \displaystyle\int\frac{dx}{1-x}=-\log|1-x| $ (the inner derivative is $\displaystyle -1 $). \[\int\frac{2\,dx}{(1-x)\left(1+x^{2}\right)}=-\log|1-x|+\frac12\log\left(1+x^{2}\right)+\tan^{-1}x+\mathrm{C} \]
Exercise 14
$\displaystyle \frac{3 x-1}{(x+2)^{2}}$
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NCERT’s answer
$\displaystyle 3 \log |x+2|+\frac{7}{x+2}+C$
A single repeated linear factor; substitute $\displaystyle t=x+2 $, so $\displaystyle x=t-2 $ and $\displaystyle dx=dt $: \[\frac{3x-1}{(x+2)^{2}}=\frac{3(t-2)-1}{t^{2}}=\frac{3t-7}{t^{2}}=\frac{3}{t}-\frac{7}{t^{2}} \] \[\int\left(\frac{3}{t}-\frac{7}{t^{2}}\right)dt=3\log|t|+\frac{7}{t}+\mathrm{C} \] Restoring $\displaystyle t=x+2 $: \[\int\frac{(3x-1)\,dx}{(x+2)^{2}}=3\log|x+2|+\frac{7}{x+2}+\mathrm{C} \]
Exercise 15
$\displaystyle \frac{1}{x^{4}-1}$
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NCERT’s answer
$\displaystyle \frac{1}{4} \log \left|\frac{x-1}{x+1}\right|-\frac{1}{2} \tan ^{-1} x+\mathrm{C}$
Factorise: $\displaystyle x^{4}-1=\left(x^{2}-1\right)\left(x^{2}+1\right) $. Split on these two blocks first, using $\displaystyle \left(x^{2}+1\right)-\left(x^{2}-1\right)=2 $: \[\frac{1}{\left(x^{2}-1\right)\left(x^{2}+1\right)}=\frac12\left(\frac{1}{x^{2}-1}-\frac{1}{x^{2}+1}\right) \] The first piece splits further, since $\displaystyle x^{2}-1=(x-1)(x+1) $: \[\frac{1}{x^{2}-1}=\frac12\left(\frac{1}{x-1}-\frac{1}{x+1}\right) \] Therefore \[\int\frac{dx}{x^{4}-1}=\frac14\int\left(\frac{1}{x-1}-\frac{1}{x+1}\right)dx-\frac12\int\frac{dx}{x^{2}+1} \] \[=\frac14\log\left|\frac{x-1}{x+1}\right|-\frac12\tan^{-1}x+\mathrm{C} \]
Exercise 16
$\displaystyle \frac{1}{x\left(x^{n}+1\right)}$ [Hint: multiply numerator and denominator by $\displaystyle x^{n-1}$ and put $\displaystyle x^{n}=t$ ]
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NCERT’s answer
$\displaystyle \frac{1}{n} \log \left|\frac{x^{n}}{x^{n}+1}\right|+\mathrm{C}$
Following the hint, multiply numerator and denominator by $\displaystyle x^{n-1} $: \[\frac{1}{x\left(x^{n}+1\right)}=\frac{x^{n-1}}{x^{n}\left(x^{n}+1\right)} \] Put $\displaystyle t=x^{n} $, so $\displaystyle dt=n\,x^{n-1}dx $, i.e. $\displaystyle x^{n-1}dx=\dfrac{dt}{n} $: \[\int\frac{dx}{x\left(x^{n}+1\right)}=\frac1n\int\frac{dt}{t(t+1)} \] Since $\displaystyle \dfrac{1}{t(t+1)}=\dfrac1t-\dfrac{1}{t+1} $, \[=\frac1n\left(\log|t|-\log|t+1|\right)+\mathrm{C}=\frac1n\log\left|\frac{t}{t+1}\right|+\mathrm{C} \] Restoring $\displaystyle t=x^{n} $: \[\int\frac{dx}{x\left(x^{n}+1\right)}=\frac1n\log\left|\frac{x^{n}}{x^{n}+1}\right|+\mathrm{C} \]
Exercise 17
$\displaystyle \frac{\cos x}{(1-\sin x)(2-\sin x)}$ [Hint : Put $\displaystyle \sin x=t$ ]
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NCERT’s answer
$\displaystyle \log \left|\frac{2-\sin x}{1-\sin x}\right|+C$
Put $\displaystyle t=\sin x $, so $\displaystyle dt=\cos x\,dx $ — the numerator is exactly $\displaystyle dt $: \[\int\frac{\cos x\,dx}{(1-\sin x)(2-\sin x)}=\int\frac{dt}{(1-t)(2-t)} \] Partial fractions: \[\frac{1}{(1-t)(2-t)}=\frac{A}{1-t}+\frac{B}{2-t},\qquad 1=A(2-t)+B(1-t) \] Put $\displaystyle t=1 $: $\displaystyle 1=A $. Put $\displaystyle t=2 $: $\displaystyle 1=-B\Rightarrow B=-1 $.Watch the signs: $\displaystyle \displaystyle\int\frac{dt}{1-t}=-\log|1-t| $ and $\displaystyle \displaystyle\int\frac{dt}{2-t}=-\log|2-t| $. \[\int\left(\frac{1}{1-t}-\frac{1}{2-t}\right)dt=-\log|1-t|+\log|2-t|+\mathrm{C} \] Restoring $\displaystyle t=\sin x $: \[=\log\left|\frac{2-\sin x}{1-\sin x}\right|+\mathrm{C} \]
Exercise 18
$\displaystyle \frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}$
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NCERT’s answer
$\displaystyle x+\frac{2}{\sqrt{3}} \tan ^{-1} \frac{x}{\sqrt{3}}-3 \tan ^{-1} \frac{x}{2}+C$
Numerator and denominator both have degree $\displaystyle 4$, so make the fraction proper first: \[\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}=\frac{x^{4}+3x^{2}+2}{x^{4}+7x^{2}+12}=1-\frac{4x^{2}+10}{\left(x^{2}+3\right)\left(x^{2}+4\right)} \] Now split the proper part, treating $\displaystyle y=x^{2} $ as the variable (do not try to split into linear factors of $\displaystyle x $ — $\displaystyle x^{2}+3 $ and $\displaystyle x^{2}+4 $ are irreducible): \[\frac{4y+10}{(y+3)(y+4)}=\frac{A}{y+3}+\frac{B}{y+4},\qquad 4y+10=A(y+4)+B(y+3) \] Put $\displaystyle y=-3 $: $\displaystyle -2=A $. Put $\displaystyle y=-4 $: $\displaystyle -6=-B\Rightarrow B=6 $.Hence the integrand equals \[1-\left(\frac{-2}{x^{2}+3}+\frac{6}{x^{2}+4}\right)=1+\frac{2}{x^{2}+3}-\frac{6}{x^{2}+4} \] Using $\displaystyle \displaystyle\int\frac{dx}{x^{2}+a^{2}}=\frac1a\tan^{-1}\frac{x}{a} $ with $\displaystyle a=\sqrt3 $ and $\displaystyle a=2 $: \[\int\frac{\left(x^{2}+1\right)\left(x^{2}+2\right)}{\left(x^{2}+3\right)\left(x^{2}+4\right)}dx=x+\frac{2}{\sqrt3}\tan^{-1}\frac{x}{\sqrt3}-3\tan^{-1}\frac{x}{2}+\mathrm{C} \]
Exercise 19
$\displaystyle \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left(\frac{x^{2}+1}{x^{2}+3}\right)+\mathrm{C}$
The numerator $\displaystyle 2x\,dx $ is the differential of $\displaystyle x^{2} $, so put $\displaystyle t=x^{2} $, $\displaystyle dt=2x\,dx $: \[\int\frac{2x\,dx}{\left(x^{2}+1\right)\left(x^{2}+3\right)}=\int\frac{dt}{(t+1)(t+3)} \] Partial fractions in $\displaystyle t $: \[\frac{1}{(t+1)(t+3)}=\frac12\left(\frac{1}{t+1}-\frac{1}{t+3}\right) \] \[\int\frac{dt}{(t+1)(t+3)}=\frac12\log\left|\frac{t+1}{t+3}\right|+\mathrm{C} \] Restoring $\displaystyle t=x^{2} $ (both $\displaystyle x^{2}+1 $ and $\displaystyle x^{2}+3 $ are positive, so the modulus may be dropped): \[=\frac12\log\frac{x^{2}+1}{x^{2}+3}+\mathrm{C} \]
Exercise 20
$\displaystyle \frac{1}{x\left(x^{4}-1\right)}$
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NCERT’s answer
$\displaystyle \frac{1}{4} \log \left|\frac{x^{4}-1}{x^{4}}\right|+\mathrm{C}$
Multiply numerator and denominator by $\displaystyle x^{3} $ so that the substitution $\displaystyle t=x^{4} $ becomes available: \[\frac{1}{x\left(x^{4}-1\right)}=\frac{x^{3}}{x^{4}\left(x^{4}-1\right)} \] Put $\displaystyle t=x^{4} $, so $\displaystyle dt=4x^{3}dx $, i.e. $\displaystyle x^{3}dx=\dfrac{dt}{4} $: \[\int\frac{dx}{x\left(x^{4}-1\right)}=\frac14\int\frac{dt}{t(t-1)} \] Since $\displaystyle \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac{1}{t} $, \[=\frac14\left(\log|t-1|-\log|t|\right)+\mathrm{C}=\frac14\log\left|\frac{t-1}{t}\right|+\mathrm{C} \] Restoring $\displaystyle t=x^{4} $: \[\int\frac{dx}{x\left(x^{4}-1\right)}=\frac14\log\left|\frac{x^{4}-1}{x^{4}}\right|+\mathrm{C} \]
Exercise 21
$\displaystyle \frac{1}{\left(e^{x}-1\right)}$ [Hint : Put $\displaystyle e^{x}=t$ ]
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NCERT’s answer
$\displaystyle \log \left(\frac{e^{x}-1}{e^{x}}\right)+\mathrm{C}$
Following the hint, put $\displaystyle t=e^{x} $. Then $\displaystyle dt=e^{x}dx=t\,dx $, so $\displaystyle dx=\dfrac{dt}{t} $ — this extra $\displaystyle \frac1t $ is the step most often dropped. \[\int\frac{dx}{e^{x}-1}=\int\frac{1}{t-1}\cdot\frac{dt}{t}=\int\frac{dt}{t(t-1)} \] Partial fractions: $\displaystyle \dfrac{1}{t(t-1)}=\dfrac{1}{t-1}-\dfrac{1}{t} $. \[\int\left(\frac{1}{t-1}-\frac{1}{t}\right)dt=\log|t-1|-\log|t|+\mathrm{C} \] Restoring $\displaystyle t=e^{x} $ and using $\displaystyle \log e^{x}=x $: \[\int\frac{dx}{e^{x}-1}=\log\left|e^{x}-1\right|-x+\mathrm{C} \]
Choose the correct answer in each of the Exercises $\displaystyle 22$ and 23.
Exercise 22
$\displaystyle \int \frac{x d x}{(x-1)(x-2)}$ equals (A) $\displaystyle \log \left|\frac{(x-1)^{2}}{x-2}\right|+\mathrm{C}$ (B) $\displaystyle \log \left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C}$ (C) $\displaystyle \log \left|\left(\frac{x-1}{x-2}\right)^{2}\right|+\mathrm{C}$ (D) $\displaystyle \log |(x-1)(x-2)|+\mathrm{C}$
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NCERT’s answer
B
Partial fractions with two distinct linear factors: \[\frac{x}{(x-1)(x-2)}=\frac{A}{x-1}+\frac{B}{x-2},\qquad x=A(x-2)+B(x-1) \] Put $\displaystyle x=1 $: $\displaystyle 1=-A\Rightarrow A=-1 $. Put $\displaystyle x=2 $: $\displaystyle 2=B $.\[\int\frac{x\,dx}{(x-1)(x-2)}=-\log|x-1|+2\log|x-2|+\mathrm{C}=\log\left|\frac{(x-2)^{2}}{x-1}\right|+\mathrm{C} \]The correct answer is (B).
Exercise 23
$\displaystyle \int \frac{d x}{x\left(x^{2}+1\right)}$ equals (A) $\displaystyle \log |x|-\frac{1}{2} \log \left(x^{2}+1\right)+C$ (B) $\displaystyle \log |x|+\frac{1}{2} \log \left(x^{2}+1\right)+C$ (C) $\displaystyle -\log |x|+\frac{1}{2} \log \left(x^{2}+1\right)+C$ (D) $\displaystyle \frac{1}{2} \log |x|+\log \left(x^{2}+1\right)+C$
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NCERT’s answer
A
$\displaystyle x^{2}+1 $ is irreducible, so it takes a linear numerator: \[\frac{1}{x\left(x^{2}+1\right)}=\frac{A}{x}+\frac{Bx+C}{x^{2}+1},\qquad 1=A\left(x^{2}+1\right)+(Bx+C)x \] Put $\displaystyle x=0 $: $\displaystyle A=1 $. Coefficient of $\displaystyle x^{2} $: $\displaystyle 0=A+B\Rightarrow B=-1 $. Coefficient of $\displaystyle x $: $\displaystyle C=0 $.\[\int\frac{dx}{x\left(x^{2}+1\right)}=\int\frac{dx}{x}-\int\frac{x\,dx}{x^{2}+1}=\log|x|-\frac12\log\left(x^{2}+1\right)+\mathrm{C} \]The correct answer is (A).

EXERCISE 7.6 1–10 (part 15 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 22.
Exercise 1
$\displaystyle x \sin x$
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NCERT’s answer
$\displaystyle -x \cos x+\sin x+\mathrm{C}$
Integration by parts: $\displaystyle \displaystyle\int u\,v\,dx = u\int v\,dx-\int\left(\frac{du}{dx}\int v\,dx\right)dx$. By the ILATE order the algebraic factor $\displaystyle x$ is the first function and $\displaystyle \sin x$ the second. \[\int x\sin x\,dx = x(-\cos x)-\int 1\cdot(-\cos x)\,dx = -x\cos x+\int\cos x\,dx\] \[= -x\cos x+\sin x+\mathrm{C}\] Final answer: $\displaystyle -x\cos x+\sin x+\mathrm{C}$.
Exercise 2
$\displaystyle x \sin 3 x$
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NCERT’s answer
$\displaystyle -\frac{x}{3} \cos 3 x+\frac{1}{9} \sin 3 x+\mathrm{C}$
Integration by parts with first function $\displaystyle x$ and second function $\displaystyle \sin 3x$; note $\displaystyle \displaystyle\int\sin 3x\,dx=-\frac{\cos 3x}{3}$ (the $\displaystyle \tfrac13$ from the chain rule is the step most often dropped). \[\int x\sin 3x\,dx = x\left(-\frac{\cos 3x}{3}\right)-\int 1\cdot\left(-\frac{\cos 3x}{3}\right)dx\] \[= -\frac{x\cos 3x}{3}+\frac13\int\cos 3x\,dx = -\frac{x\cos 3x}{3}+\frac{\sin 3x}{9}+\mathrm{C}\] Final answer: $\displaystyle -\dfrac{x\cos 3x}{3}+\dfrac{\sin 3x}{9}+\mathrm{C}$.
Exercise 3
$\displaystyle x^{2} e^{x}$
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NCERT’s answer
$\displaystyle e^{x}\left(x^{2}-2 x+2\right)+\mathrm{C}$
Integration by parts, first function $\displaystyle x^{2}$, second function $\displaystyle e^{x}$. The parts have to be applied twice, because one application only lowers the power of $\displaystyle x$ by one. \[\int x^{2}e^{x}\,dx = x^{2}e^{x}-\int 2x\,e^{x}\,dx\] For the new integral, parts again with first function $\displaystyle x$: \[\int x e^{x}\,dx = x e^{x}-\int e^{x}\,dx = x e^{x}-e^{x}\] Hence \[\int x^{2}e^{x}\,dx = x^{2}e^{x}-2\left(x e^{x}-e^{x}\right)+\mathrm{C} = e^{x}\left(x^{2}-2x+2\right)+\mathrm{C}\] Final answer: $\displaystyle e^{x}\left(x^{2}-2x+2\right)+\mathrm{C}$.
Exercise 4
$\displaystyle x \log x$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2} \log x-\frac{x^{2}}{4}+\mathrm{C}$
Here the logarithm has no elementary antiderivative but an easy derivative, so by ILATE $\displaystyle \log x$ is the FIRST function and $\displaystyle x$ the second. \[\int x\log x\,dx = \log x\cdot\frac{x^{2}}{2}-\int\frac{1}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\log x-\frac12\int x\,dx\] \[= \frac{x^{2}}{2}\log x-\frac{x^{2}}{4}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{x^{2}}{2}\log x-\dfrac{x^{2}}{4}+\mathrm{C}$, valid for $\displaystyle x>0$.
Exercise 5
$\displaystyle x \log 2 x$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2} \log 2 x-\frac{x^{2}}{4}+\mathrm{C}$
Take $\displaystyle \log 2x$ as the first function and $\displaystyle x$ as the second. The key point: $\displaystyle \dfrac{d}{dx}\log 2x=\dfrac{1}{2x}\cdot 2=\dfrac{1}{x}$ — the same derivative as $\displaystyle \log x$, since $\displaystyle \log 2x=\log 2+\log x$. \[\int x\log 2x\,dx = \log 2x\cdot\frac{x^{2}}{2}-\int\frac{1}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\log 2x-\frac12\int x\,dx\] \[= \frac{x^{2}}{2}\log 2x-\frac{x^{2}}{4}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{x^{2}}{2}\log 2x-\dfrac{x^{2}}{4}+\mathrm{C}$, valid for $\displaystyle x>0$.
Exercise 6
$\displaystyle x^{2} \log x$
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NCERT’s answer
$\displaystyle \frac{x^{3}}{3} \log x-\frac{x^{3}}{9}+\mathrm{C}$
By ILATE, $\displaystyle \log x$ is the first function and $\displaystyle x^{2}$ the second. \[\int x^{2}\log x\,dx = \log x\cdot\frac{x^{3}}{3}-\int\frac{1}{x}\cdot\frac{x^{3}}{3}\,dx = \frac{x^{3}}{3}\log x-\frac13\int x^{2}\,dx\] \[= \frac{x^{3}}{3}\log x-\frac{x^{3}}{9}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{x^{3}}{3}\log x-\dfrac{x^{3}}{9}+\mathrm{C}$, valid for $\displaystyle x>0$.
Exercise 7
$\displaystyle x \sin ^{-1} x$
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NCERT’s answer
$\displaystyle \frac{1}{4}\left(2 x^{2}-1\right) \sin ^{-1} x+\frac{x \sqrt{1-x^{2}}}{4}+\mathrm{C}$
By ILATE the inverse-trigonometric factor $\displaystyle \sin^{-1}x$ is the first function, $\displaystyle x$ the second, and $\displaystyle \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}$. \[\int x\sin^{-1}x\,dx = \frac{x^{2}}{2}\sin^{-1}x-\frac12\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx\] For the remaining integral write $\displaystyle x^{2}=-\left(1-x^{2}\right)+1$, which splits it into two standard forms: \[\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \int\frac{dx}{\sqrt{1-x^{2}}}-\int\sqrt{1-x^{2}}\,dx = \sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^{2}}+\frac12\sin^{-1}x\right]\] \[= \frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\] Substituting back, \[\int x\sin^{-1}x\,dx = \frac{x^{2}}{2}\sin^{-1}x-\frac14\sin^{-1}x+\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{2x^{2}-1}{4}\sin^{-1}x+\dfrac{x\sqrt{1-x^{2}}}{4}+\mathrm{C}$, for $\displaystyle -1\le x\le 1$.
Exercise 8
$\displaystyle x \tan ^{-1} x$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2} \tan ^{-1} x-\frac{x}{2}+\frac{1}{2} \tan ^{-1} x+\mathrm{C}$
By ILATE, $\displaystyle \tan^{-1}x$ is the first function and $\displaystyle x$ the second; $\displaystyle \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}$. \[\int x\tan^{-1}x\,dx = \frac{x^{2}}{2}\tan^{-1}x-\frac12\int\frac{x^{2}}{1+x^{2}}\,dx\] The fraction is improper, so divide before integrating: $\displaystyle \dfrac{x^{2}}{1+x^{2}}=1-\dfrac{1}{1+x^{2}}$. \[\int\frac{x^{2}}{1+x^{2}}\,dx = \int\left(1-\frac{1}{1+x^{2}}\right)dx = x-\tan^{-1}x\] Hence \[\int x\tan^{-1}x\,dx = \frac{x^{2}}{2}\tan^{-1}x-\frac{x}{2}+\frac12\tan^{-1}x+\mathrm{C}\] Final answer: $\displaystyle \dfrac{x^{2}+1}{2}\tan^{-1}x-\dfrac{x}{2}+\mathrm{C}$.
Exercise 9
$\displaystyle x \cos ^{-1} x$
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NCERT’s answer
$\displaystyle \left(2 x^{2}-1\right) \frac{\cos ^{-1} x}{4}-\frac{x}{4} \sqrt{1-x^{2}}+\mathrm{C}$
By ILATE, $\displaystyle \cos^{-1}x$ is the first function and $\displaystyle x$ the second. The sign is the trap here: $\displaystyle \dfrac{d}{dx}\cos^{-1}x=-\dfrac{1}{\sqrt{1-x^{2}}}$. \[\int x\cos^{-1}x\,dx = \frac{x^{2}}{2}\cos^{-1}x-\int\left(-\frac{1}{\sqrt{1-x^{2}}}\right)\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}\cos^{-1}x+\frac12\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx\] Writing $\displaystyle x^{2}=1-\left(1-x^{2}\right)$, \[\int\frac{x^{2}}{\sqrt{1-x^{2}}}\,dx = \sin^{-1}x-\left[\frac{x}{2}\sqrt{1-x^{2}}+\frac12\sin^{-1}x\right] = \frac12\sin^{-1}x-\frac{x}{2}\sqrt{1-x^{2}}\] So \[\int x\cos^{-1}x\,dx = \frac{x^{2}}{2}\cos^{-1}x+\frac14\sin^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Using $\displaystyle \sin^{-1}x=\dfrac{\pi}{2}-\cos^{-1}x$ and absorbing the constant $\displaystyle \dfrac{\pi}{8}$ into $\displaystyle \mathrm{C}$: \[= \frac{2x^{2}-1}{4}\cos^{-1}x-\frac{x}{4}\sqrt{1-x^{2}}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{2x^{2}-1}{4}\cos^{-1}x-\dfrac{x\sqrt{1-x^{2}}}{4}+\mathrm{C}$, for $\displaystyle -1\le x\le 1$.
Exercise 10
$\displaystyle \left(\sin ^{-1} x\right)^{2}$
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NCERT’s answer
$\displaystyle \left(\sin ^{-1} x\right)^{2} x+2 \sqrt{1-x^{2}} \sin ^{-1} x-2 x+\mathrm{C}$
Substitute $\displaystyle x=\sin\theta$, so $\displaystyle \theta=\sin^{-1}x$ with $\displaystyle \theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ and $\displaystyle dx=\cos\theta\,d\theta$. \[I=\int\left(\sin^{-1}x\right)^{2}dx = \int\theta^{2}\cos\theta\,d\theta\] Parts with first function $\displaystyle \theta^{2}$: \[\int\theta^{2}\cos\theta\,d\theta = \theta^{2}\sin\theta-\int 2\theta\sin\theta\,d\theta\] Parts again on $\displaystyle \displaystyle\int\theta\sin\theta\,d\theta=-\theta\cos\theta+\sin\theta$, so \[I = \theta^{2}\sin\theta-2\left(-\theta\cos\theta+\sin\theta\right)+\mathrm{C} = \theta^{2}\sin\theta+2\theta\cos\theta-2\sin\theta+\mathrm{C}\] On $\displaystyle \left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$ the cosine is non-negative, so $\displaystyle \cos\theta=+\sqrt{1-x^{2}}$ (this positive root is the step to be careful about). Putting $\displaystyle \theta=\sin^{-1}x$, $\displaystyle \sin\theta=x$: \[I = x\left(\sin^{-1}x\right)^{2}+2\sqrt{1-x^{2}}\,\sin^{-1}x-2x+\mathrm{C}\] Final answer: $\displaystyle x\left(\sin^{-1}x\right)^{2}+2\sqrt{1-x^{2}}\,\sin^{-1}x-2x+\mathrm{C}$, for $\displaystyle -1\le x\le 1$.

EXERCISE 7.6 11–20 (part 16 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 22.
Exercise 11
$\displaystyle \frac{x \cos ^{-1} x}{\sqrt{1-x^{2}}}$
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NCERT’s answer
$\displaystyle -\sqrt{1-x^{2}} \cos ^{-1} x+x+\mathrm{C}$
Substitute $\displaystyle \theta=\cos^{-1}x$, so $\displaystyle x=\cos\theta$, $\displaystyle dx=-\sin\theta\,d\theta$, and $\displaystyle \sqrt{1-x^{2}}=\sin\theta$ (positive, since $\displaystyle \theta\in[0,\pi]$). \[I=\int\frac{x\cos^{-1}x}{\sqrt{1-x^{2}}}\,dx = \int\frac{\cos\theta\cdot\theta}{\sin\theta}\left(-\sin\theta\right)d\theta = -\int\theta\cos\theta\,d\theta\] By parts, first function $\displaystyle \theta$: \[\int\theta\cos\theta\,d\theta = \theta\sin\theta-\int\sin\theta\,d\theta = \theta\sin\theta+\cos\theta\] So $\displaystyle I=-\theta\sin\theta-\cos\theta+\mathrm{C}$. Restoring $\displaystyle \theta=\cos^{-1}x$, $\displaystyle \sin\theta=\sqrt{1-x^{2}}$, $\displaystyle \cos\theta=x$: \[I = -\sqrt{1-x^{2}}\,\cos^{-1}x-x+\mathrm{C}\] Final answer: $\displaystyle -\sqrt{1-x^{2}}\,\cos^{-1}x-x+\mathrm{C}$, for $\displaystyle -1<x<1$.
Exercise 12
$\displaystyle x \sec ^{2} x$
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NCERT’s answer
$\displaystyle x \tan x+\log |\cos x|+\mathrm{C}$
Parts with first function $\displaystyle x$ and second function $\displaystyle \sec^{2}x$, whose integral is $\displaystyle \tan x$. \[\int x\sec^{2}x\,dx = x\tan x-\int 1\cdot\tan x\,dx\] Using the standard result $\displaystyle \displaystyle\int\tan x\,dx=-\log|\cos x|+\mathrm{C}$, \[= x\tan x+\log|\cos x|+\mathrm{C}\] Final answer: $\displaystyle x\tan x+\log|\cos x|+\mathrm{C}$ (equivalently $\displaystyle x\tan x-\log|\sec x|+\mathrm{C}$), for $\displaystyle x\neq(2n+1)\dfrac{\pi}{2}$.
Exercise 13
$\displaystyle \tan ^{-1} x$
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NCERT’s answer
$\displaystyle x \tan ^{-1} x-\frac{1}{2} \log \left(1+x^{2}\right)+C$
A single inverse function is integrated by parts against the second function $\displaystyle 1$. \[\int\tan^{-1}x\,dx = \int\left(\tan^{-1}x\right)\cdot 1\,dx = \tan^{-1}x\cdot x-\int\frac{1}{1+x^{2}}\cdot x\,dx\] For the last integral put $\displaystyle t=1+x^{2}$, $\displaystyle dt=2x\,dx$: \[\int\frac{x}{1+x^{2}}\,dx = \frac12\int\frac{dt}{t} = \frac12\log\left(1+x^{2}\right)\] Hence \[\int\tan^{-1}x\,dx = x\tan^{-1}x-\frac12\log\left(1+x^{2}\right)+\mathrm{C}\] Final answer: $\displaystyle x\tan^{-1}x-\dfrac12\log\left(1+x^{2}\right)+\mathrm{C}$.
Exercise 14
$\displaystyle x(\log x)^{2}$
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NCERT’s answer
$\displaystyle \frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2} \log x+\frac{x^{2}}{4}+\mathrm{C}$
By ILATE, $\displaystyle (\log x)^{2}$ is the first function and $\displaystyle x$ the second; $\displaystyle \dfrac{d}{dx}(\log x)^{2}=\dfrac{2\log x}{x}$ by the chain rule. \[\int x(\log x)^{2}dx = (\log x)^{2}\frac{x^{2}}{2}-\int\frac{2\log x}{x}\cdot\frac{x^{2}}{2}\,dx = \frac{x^{2}}{2}(\log x)^{2}-\int x\log x\,dx\] The leftover integral is Question $\displaystyle 4$: $\displaystyle \displaystyle\int x\log x\,dx=\frac{x^{2}}{2}\log x-\frac{x^{2}}{4}$. Therefore \[\int x(\log x)^{2}dx = \frac{x^{2}}{2}(\log x)^{2}-\frac{x^{2}}{2}\log x+\frac{x^{2}}{4}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{x^{2}}{2}(\log x)^{2}-\dfrac{x^{2}}{2}\log x+\dfrac{x^{2}}{4}+\mathrm{C}$, valid for $\displaystyle x>0$.
Exercise 15
$\displaystyle \left(x^{2}+1\right) \log x$
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NCERT’s answer
$\displaystyle \left(\frac{x^{3}}{3}+x\right) \log x-\frac{x^{3}}{9}-x+\mathrm{C}$
Take $\displaystyle \log x$ as the first function and $\displaystyle x^{2}+1$ as the second, so the second must be integrated first: $\displaystyle \displaystyle\int\left(x^{2}+1\right)dx=\frac{x^{3}}{3}+x$. \[\int\left(x^{2}+1\right)\log x\,dx = \log x\left(\frac{x^{3}}{3}+x\right)-\int\frac{1}{x}\left(\frac{x^{3}}{3}+x\right)dx\] \[= \left(\frac{x^{3}}{3}+x\right)\log x-\int\left(\frac{x^{2}}{3}+1\right)dx = \left(\frac{x^{3}}{3}+x\right)\log x-\frac{x^{3}}{9}-x+\mathrm{C}\] Final answer: $\displaystyle \left(\dfrac{x^{3}}{3}+x\right)\log x-\dfrac{x^{3}}{9}-x+\mathrm{C}$, valid for $\displaystyle x>0$.
Exercise 16
$\displaystyle e^{x}(\sin x+\cos x)$
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NCERT’s answer
$\displaystyle e^{x} \sin x+\mathrm{C}$
Use the standard result $\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx = e^{x}f(x)+\mathrm{C}$. Here the bracket is $\displaystyle \sin x+\cos x$; taking $\displaystyle f(x)=\sin x$ gives $\displaystyle f'(x)=\cos x$, so the integrand is exactly $\displaystyle e^{x}\left[f(x)+f'(x)\right]$. \[\int e^{x}\left(\sin x+\cos x\right)dx = e^{x}\sin x+\mathrm{C}\] Final answer: $\displaystyle e^{x}\sin x+\mathrm{C}$.
Exercise 17
$\displaystyle \frac{x e^{x}}{(1+x)^{2}}$
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NCERT’s answer
$\displaystyle \frac{e^{x}}{1+x}+\mathrm{C}$
Aim for the form $\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}$, so split the rational factor by writing the numerator in terms of $\displaystyle 1+x$: \[\frac{x}{(1+x)^{2}} = \frac{(1+x)-1}{(1+x)^{2}} = \frac{1}{1+x}-\frac{1}{(1+x)^{2}}\] With $\displaystyle f(x)=\dfrac{1}{1+x}$ we get $\displaystyle f'(x)=-\dfrac{1}{(1+x)^{2}}$, so the integrand is $\displaystyle e^{x}\left[f(x)+f'(x)\right]$. \[\int\frac{x e^{x}}{(1+x)^{2}}\,dx = \frac{e^{x}}{1+x}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{e^{x}}{1+x}+\mathrm{C}$, for $\displaystyle x\neq-1$.
Exercise 18
$\displaystyle e^{x}\left(\frac{1+\sin x}{1+\cos x}\right)$
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NCERT’s answer
$\displaystyle e^{x} \tan \frac{x}{2}+\mathrm{C}$
Convert to half angles using $\displaystyle 1+\cos x = 2\cos^{2}\dfrac{x}{2}$ and $\displaystyle \sin x = 2\sin\dfrac{x}{2}\cos\dfrac{x}{2}$: \[\frac{1+\sin x}{1+\cos x} = \frac{1+2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\cos^{2}\dfrac{x}{2}} = \frac12\sec^{2}\frac{x}{2}+\tan\frac{x}{2}\] So the integrand is $\displaystyle e^{x}\left[f(x)+f'(x)\right]$ with $\displaystyle f(x)=\tan\dfrac{x}{2}$, since $\displaystyle f'(x)=\dfrac12\sec^{2}\dfrac{x}{2}$. \[\int e^{x}\left(\frac{1+\sin x}{1+\cos x}\right)dx = e^{x}\tan\frac{x}{2}+\mathrm{C}\] Final answer: $\displaystyle e^{x}\tan\dfrac{x}{2}+\mathrm{C}$, for $\displaystyle x\neq(2n+1)\pi$.
Exercise 19
$\displaystyle e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)$
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NCERT’s answer
$\displaystyle \frac{e^{x}}{x}+\mathrm{C}$
Match the pattern $\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}$. Taking $\displaystyle f(x)=\dfrac{1}{x}$ gives $\displaystyle f'(x)=-\dfrac{1}{x^{2}}$, and the bracket $\displaystyle \dfrac{1}{x}-\dfrac{1}{x^{2}}$ is exactly $\displaystyle f(x)+f'(x)$. \[\int e^{x}\left(\frac{1}{x}-\frac{1}{x^{2}}\right)dx = \frac{e^{x}}{x}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{e^{x}}{x}+\mathrm{C}$, for $\displaystyle x\neq 0$.
Exercise 20
$\displaystyle \frac{(x-3) e^{x}}{(x-1)^{3}}$
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NCERT’s answer
$\displaystyle \frac{e^{x}}{(x-1)^{2}}+\mathrm{C}$
Rewrite the numerator in terms of $\displaystyle x-1$ so the $\displaystyle e^{x}\left[f(x)+f'(x)\right]$ pattern appears: $\displaystyle x-3=(x-1)-2$, hence \[\frac{x-3}{(x-1)^{3}} = \frac{(x-1)-2}{(x-1)^{3}} = \frac{1}{(x-1)^{2}}-\frac{2}{(x-1)^{3}}\] With $\displaystyle f(x)=\dfrac{1}{(x-1)^{2}}$ we have $\displaystyle f'(x)=-\dfrac{2}{(x-1)^{3}}$, so the integrand is $\displaystyle e^{x}\left[f(x)+f'(x)\right]$. \[\int\frac{(x-3)e^{x}}{(x-1)^{3}}\,dx = \frac{e^{x}}{(x-1)^{2}}+\mathrm{C}\] Final answer: $\displaystyle \dfrac{e^{x}}{(x-1)^{2}}+\mathrm{C}$, for $\displaystyle x\neq 1$.

EXERCISE 7.6 21–24 (part 17 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 22.
Exercise 21
$\displaystyle e^{2 x} \sin x$
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NCERT’s answer
$\displaystyle \frac{e^{2 x}}{5}(2 \sin x-\cos x)+C$
Let $\displaystyle I=\displaystyle\int e^{2x}\sin x\,dx$. Apply parts twice; the original integral reappears and is then solved for algebraically. First, with $\displaystyle \sin x$ as the first function and $\displaystyle e^{2x}$ as the second: \[I = \frac{e^{2x}}{2}\sin x-\frac12\int e^{2x}\cos x\,dx\] Apply parts to the new integral, again with the trigonometric factor first: \[\int e^{2x}\cos x\,dx = \frac{e^{2x}}{2}\cos x+\frac12\int e^{2x}\sin x\,dx = \frac{e^{2x}}{2}\cos x+\frac{I}{2}\] Substituting back: \[I = \frac{e^{2x}}{2}\sin x-\frac{e^{2x}}{4}\cos x-\frac{I}{4}\quad\Rightarrow\quad \frac{5I}{4} = \frac{e^{2x}}{4}\left(2\sin x-\cos x\right)\] \[I = \frac{e^{2x}}{5}\left(2\sin x-\cos x\right)+\mathrm{C}\] Final answer: $\displaystyle \dfrac{e^{2x}}{5}\left(2\sin x-\cos x\right)+\mathrm{C}$.
Exercise 22
$\displaystyle \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$
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NCERT’s answer
$\displaystyle 2 x \tan ^{-1} x-\log \left(1+x^{2}\right)+\mathrm{C}$
First simplify the inverse function. Put $\displaystyle x=\tan\theta$, i.e. $\displaystyle \theta=\tan^{-1}x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$. Then \[\frac{2x}{1+x^{2}} = \frac{2\tan\theta}{1+\tan^{2}\theta} = \sin 2\theta\] Now $\displaystyle \sin^{-1}(\sin 2\theta)=2\theta$ only when $\displaystyle 2\theta\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$, i.e. $\displaystyle \theta\in\left[-\dfrac{\pi}{4},\dfrac{\pi}{4}\right]$, i.e. $\displaystyle -1\le x\le 1$. This range check is the step that decides the answer; taking it as $\displaystyle 2\tan^{-1}x$ for all $\displaystyle x$ is wrong. For $\displaystyle -1\le x\le 1$, \[\int\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)dx = 2\int\tan^{-1}x\,dx = 2\left[x\tan^{-1}x-\frac12\log\left(1+x^{2}\right)\right]+\mathrm{C}\] using Question 13. Final answer: $\displaystyle 2x\tan^{-1}x-\log\left(1+x^{2}\right)+\mathrm{C}$, for $\displaystyle -1\le x\le 1$. (Outside that range the identity changes to $\displaystyle \pi-2\tan^{-1}x$ for $\displaystyle x>1$ and $\displaystyle -\pi-2\tan^{-1}x$ for $\displaystyle x<-1$, giving $\displaystyle \pi x-2x\tan^{-1}x+\log\left(1+x^{2}\right)+\mathrm{C}$ and $\displaystyle -\pi x-2x\tan^{-1}x+\log\left(1+x^{2}\right)+\mathrm{C}$ respectively.)
Choose the correct answer in Exercises $\displaystyle 23$ and 24.
Exercise 23
$\displaystyle \int x^{2} e^{x^{3}} d x$ equals (A) $\displaystyle \frac{1}{3} e^{x^{3}}+\mathrm{C}$ (B) $\displaystyle \frac{1}{3} e^{x^{2}}+\mathrm{C}$ (C) $\displaystyle \frac{1}{2} e^{x^{3}}+\mathrm{C}$ (D) $\displaystyle \frac{1}{2} e^{x^{2}}+\mathrm{C}$
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NCERT’s answer
A
This is a substitution, not integration by parts: the factor $\displaystyle x^{2}$ is (up to a constant) the derivative of the exponent $\displaystyle x^{3}$. Put $\displaystyle t=x^{3}$, so $\displaystyle dt=3x^{2}dx$, i.e. $\displaystyle x^{2}dx=\dfrac{dt}{3}$. \[\int x^{2}e^{x^{3}}dx = \frac13\int e^{t}\,dt = \frac13 e^{t}+\mathrm{C} = \frac13 e^{x^{3}}+\mathrm{C}\] Check by differentiating: $\displaystyle \dfrac{d}{dx}\left(\dfrac13 e^{x^{3}}\right)=\dfrac13 e^{x^{3}}\cdot 3x^{2}=x^{2}e^{x^{3}}$. Final answer: option (A), $\displaystyle \dfrac13 e^{x^{3}}+\mathrm{C}$.
Exercise 24
$\displaystyle \int e^{x} \sec x(1+\tan x) d x$ equals (A) $\displaystyle e^{x} \cos x+\mathrm{C}$ (B) $\displaystyle e^{x} \sec x+\mathrm{C}$ (C) $\displaystyle e^{x} \sin x+\mathrm{C}$ (D) $\displaystyle e^{x} \tan x+\mathrm{C}$
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NCERT’s answer
B
Expand the bracket to expose the standard form $\displaystyle \displaystyle\int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}$: \[\int e^{x}\sec x(1+\tan x)\,dx = \int e^{x}\left(\sec x+\sec x\tan x\right)dx\] With $\displaystyle f(x)=\sec x$, $\displaystyle f'(x)=\sec x\tan x$, so the integrand is exactly $\displaystyle e^{x}\left[f(x)+f'(x)\right]$. \[= e^{x}\sec x+\mathrm{C}\] Check by differentiating: $\displaystyle \dfrac{d}{dx}\left(e^{x}\sec x\right)=e^{x}\sec x+e^{x}\sec x\tan x=e^{x}\sec x(1+\tan x)$. Final answer: option (B), $\displaystyle e^{x}\sec x+\mathrm{C}$.

EXERCISE 7.7 1–11 (part 18 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to $\displaystyle 9$ .
Exercise 1
$\displaystyle \sqrt{4-x^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{2} x \sqrt{4-x^{2}}+2 \sin ^{-1} \frac{x}{2}+C$
The integrand has the form $\displaystyle \sqrt{a^{2}-x^{2}}$ with $\displaystyle a=2$; it is real only for $\displaystyle -2\le x\le 2$.Derive the standard result by integration by parts, taking $\displaystyle \sqrt{4-x^{2}}$ as the first function and $\displaystyle 1$ as the second: \[I=\int\sqrt{4-x^{2}}\,dx=x\sqrt{4-x^{2}}-\int x\cdot\frac{-x}{\sqrt{4-x^{2}}}\,dx=x\sqrt{4-x^{2}}+\int\frac{x^{2}}{\sqrt{4-x^{2}}}\,dx.\]The step to get right is rewriting the numerator as $\displaystyle x^{2}=-\left(4-x^{2}\right)+4$, which sends the last integral back to $\displaystyle I$: \[\int\frac{x^{2}}{\sqrt{4-x^{2}}}\,dx=-\int\sqrt{4-x^{2}}\,dx+4\int\frac{dx}{\sqrt{4-x^{2}}}=-I+4\sin^{-1}\frac{x}{2}.\]So $\displaystyle I=x\sqrt{4-x^{2}}-I+4\sin^{-1}\dfrac{x}{2}$, i.e. $\displaystyle 2I=x\sqrt{4-x^{2}}+4\sin^{-1}\dfrac{x}{2}$.\[\int\sqrt{4-x^{2}}\,dx=\frac{x}{2}\sqrt{4-x^{2}}+2\sin^{-1}\frac{x}{2}+\mathrm{C},\qquad -2\le x\le 2.\]The same argument with $\displaystyle a$ in place of $\displaystyle 2$ gives the formula used throughout this exercise: \[\int\sqrt{a^{2}-x^{2}}\,dx=\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}+\mathrm{C}.\]
Exercise 2
$\displaystyle \sqrt{1-4 x^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{4} \sin ^{-1} 2 x+\frac{1}{2} x \sqrt{1-4 x^{2}}+\mathrm{C}$
First make the coefficient of $\displaystyle x^{2}$ equal to $\displaystyle 1$ inside the root by taking the $\displaystyle 4$ out — this is the step most often dropped: \[\sqrt{1-4x^{2}}=\sqrt{4\left(\frac{1}{4}-x^{2}\right)}=2\sqrt{\left(\frac{1}{2}\right)^{2}-x^{2}}.\]Now apply $\displaystyle \int\sqrt{a^{2}-x^{2}}\,dx=\dfrac{x}{2}\sqrt{a^{2}-x^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{x}{a}+\mathrm{C}$ with $\displaystyle a=\dfrac{1}{2}$: \[\int\sqrt{1-4x^{2}}\,dx=2\left[\frac{x}{2}\sqrt{\frac{1}{4}-x^{2}}+\frac{1}{8}\sin^{-1}(2x)\right]=x\sqrt{\frac{1}{4}-x^{2}}+\frac{1}{4}\sin^{-1}(2x).\]Since $\displaystyle \sqrt{\frac{1}{4}-x^{2}}=\frac{1}{2}\sqrt{1-4x^{2}}$, \[\int\sqrt{1-4x^{2}}\,dx=\frac{x}{2}\sqrt{1-4x^{2}}+\frac{1}{4}\sin^{-1}(2x)+\mathrm{C},\qquad -\frac{1}{2}\le x\le\frac{1}{2}.\]
Exercise 3
$\displaystyle \sqrt{x^{2}+4 x+6}$
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NCERT’s answer
$\displaystyle \frac{(x+2)}{2} \sqrt{x^{2}+4 x+6}+\log \left|x+2+\sqrt{x^{2}+4 x+6}\right|+\mathrm{C}$
Complete the square: \[x^{2}+4x+6=\left(x^{2}+4x+4\right)+2=(x+2)^{2}+\left(\sqrt{2}\right)^{2}.\]Put $\displaystyle t=x+2$, so $\displaystyle dt=dx$; the integral becomes $\displaystyle \displaystyle\int\sqrt{t^{2}+a^{2}}\,dt$ with $\displaystyle a=\sqrt{2}$. Using \[\int\sqrt{t^{2}+a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}+a^{2}}+\frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}+a^{2}}\right|+\mathrm{C},\] we get, with $\displaystyle \dfrac{a^{2}}{2}=1$, \[\int\sqrt{x^{2}+4x+6}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x+6}+\log\left|x+2+\sqrt{x^{2}+4x+6}\right|+\mathrm{C}.\]Since $\displaystyle (x+2)^{2}+2>0$ for every $\displaystyle x$, this holds for all real $\displaystyle x$.
Exercise 4
$\displaystyle \sqrt{x^{2}+4 x+1}$
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NCERT’s answer
$\displaystyle \frac{(x+2)}{2} \sqrt{x^{2}+4 x+1}-\frac{3}{2} \log \left|x+2+\sqrt{x^{2}+4 x+1}\right|+\mathrm{C}$
Complete the square: \[x^{2}+4x+1=\left(x^{2}+4x+4\right)-3=(x+2)^{2}-\left(\sqrt{3}\right)^{2}.\]This is the $\displaystyle \sqrt{t^{2}-a^{2}}$ type (note the minus sign before the log, unlike Q3), with $\displaystyle t=x+2,\ dt=dx,\ a=\sqrt{3}$: \[\int\sqrt{t^{2}-a^{2}}\,dt=\frac{t}{2}\sqrt{t^{2}-a^{2}}-\frac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}.\]Hence \[\int\sqrt{x^{2}+4x+1}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x+1}-\frac{3}{2}\log\left|x+2+\sqrt{x^{2}+4x+1}\right|+\mathrm{C},\] valid where the radicand is non-negative, i.e. $\displaystyle |x+2|\ge\sqrt{3}$.
Exercise 5
$\displaystyle \sqrt{1-4 x-x^{2}}$
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NCERT’s answer
$\displaystyle \frac{5}{2} \sin ^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+\frac{x+2}{2} \sqrt{1-4 x-x^{2}}+\mathrm{C}$
Take out the sign of $\displaystyle x^{2}$ before completing the square — that is where the sign error usually creeps in: \[1-4x-x^{2}=1-\left(x^{2}+4x\right)=1-\left[(x+2)^{2}-4\right]=5-(x+2)^{2}=\left(\sqrt{5}\right)^{2}-(x+2)^{2}.\]With $\displaystyle t=x+2,\ dt=dx,\ a=\sqrt{5}$, apply $\displaystyle \int\sqrt{a^{2}-t^{2}}\,dt=\dfrac{t}{2}\sqrt{a^{2}-t^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{t}{a}+\mathrm{C}$: \[\int\sqrt{1-4x-x^{2}}\,dx=\frac{x+2}{2}\sqrt{1-4x-x^{2}}+\frac{5}{2}\sin^{-1}\left(\frac{x+2}{\sqrt{5}}\right)+\mathrm{C},\] valid on $\displaystyle |x+2|\le\sqrt{5}$, i.e. $\displaystyle -2-\sqrt{5}\le x\le -2+\sqrt{5}$.
Exercise 6
$\displaystyle \sqrt{x^{2}+4 x-5}$
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NCERT’s answer
$\displaystyle \frac{(x+2)}{2} \sqrt{x^{2}+4 x-5}-\frac{9}{2} \log \left|x+2+\sqrt{x^{2}+4 x-5}\right|+\mathrm{C}$
Complete the square: \[x^{2}+4x-5=\left(x^{2}+4x+4\right)-9=(x+2)^{2}-3^{2}.\]With $\displaystyle t=x+2,\ dt=dx,\ a=3$, use $\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}$: \[\int\sqrt{x^{2}+4x-5}\,dx=\frac{x+2}{2}\sqrt{x^{2}+4x-5}-\frac{9}{2}\log\left|x+2+\sqrt{x^{2}+4x-5}\right|+\mathrm{C}.\]The radicand $\displaystyle (x+5)(x-1)$ is non-negative only for $\displaystyle x\le -5$ or $\displaystyle x\ge 1$, so the answer holds on those intervals.
Exercise 7
$\displaystyle \sqrt{1+3 x-x^{2}}$
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NCERT’s answer
$\displaystyle \frac{(2 x-3)}{4} \sqrt{1+3 x-x^{2}}+\frac{13}{8} \sin ^{-1}\left(\frac{2 x-3}{\sqrt{13}}\right)+\mathrm{C}$
Complete the square after factoring out $\displaystyle -1$ from the $\displaystyle x$-terms: \[1+3x-x^{2}=1-\left(x^{2}-3x\right)=1-\left[\left(x-\frac{3}{2}\right)^{2}-\frac{9}{4}\right]=\frac{13}{4}-\left(x-\frac{3}{2}\right)^{2}.\]With $\displaystyle t=x-\dfrac{3}{2},\ dt=dx,\ a=\dfrac{\sqrt{13}}{2}$, apply $\displaystyle \int\sqrt{a^{2}-t^{2}}\,dt=\dfrac{t}{2}\sqrt{a^{2}-t^{2}}+\dfrac{a^{2}}{2}\sin^{-1}\dfrac{t}{a}+\mathrm{C}$. Here $\displaystyle \dfrac{a^{2}}{2}=\dfrac{13}{8}$ and $\displaystyle \dfrac{t}{a}=\dfrac{2x-3}{\sqrt{13}}$: \[\int\sqrt{1+3x-x^{2}}\,dx=\frac{2x-3}{4}\sqrt{1+3x-x^{2}}+\frac{13}{8}\sin^{-1}\left(\frac{2x-3}{\sqrt{13}}\right)+\mathrm{C},\] valid where $\displaystyle (2x-3)^{2}\le 13$, i.e. $\displaystyle \dfrac{3-\sqrt{13}}{2}\le x\le\dfrac{3+\sqrt{13}}{2}$.
Exercise 8
$\displaystyle \sqrt{x^{2}+3 x}$
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NCERT’s answer
$\displaystyle \frac{2 x+3}{4} \sqrt{x^{2}+3 x}-\frac{9}{8} \log \left|x+\frac{3}{2}+\sqrt{x^{2}+3 x}\right|+\mathrm{C}$
Complete the square: \[x^{2}+3x=\left(x+\frac{3}{2}\right)^{2}-\frac{9}{4}=\left(x+\frac{3}{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}.\]With $\displaystyle t=x+\dfrac{3}{2},\ dt=dx,\ a=\dfrac{3}{2}$, use $\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}$, where $\displaystyle \dfrac{a^{2}}{2}=\dfrac{9}{8}$: \[\int\sqrt{x^{2}+3x}\,dx=\frac{2x+3}{4}\sqrt{x^{2}+3x}-\frac{9}{8}\log\left|x+\frac{3}{2}+\sqrt{x^{2}+3x}\right|+\mathrm{C}.\]The radicand $\displaystyle x(x+3)$ is non-negative only for $\displaystyle x\le -3$ or $\displaystyle x\ge 0$, so the result holds there.
Exercise 9
$\displaystyle \sqrt{1+\frac{x^{2}}{9}}$
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NCERT’s answer
$\displaystyle \frac{x}{6} \sqrt{x^{2}+9}+\frac{3}{2} \log \left|x+\sqrt{x^{2}+9}\right|+\mathrm{C}$
Bring the fraction out of the root first: \[\sqrt{1+\frac{x^{2}}{9}}=\sqrt{\frac{9+x^{2}}{9}}=\frac{1}{3}\sqrt{x^{2}+3^{2}}.\]Apply $\displaystyle \int\sqrt{x^{2}+a^{2}}\,dx=\dfrac{x}{2}\sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2}\log\left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}$ with $\displaystyle a=3$: \[\int\sqrt{1+\frac{x^{2}}{9}}\,dx=\frac{1}{3}\left[\frac{x}{2}\sqrt{x^{2}+9}+\frac{9}{2}\log\left|x+\sqrt{x^{2}+9}\right|\right]+\mathrm{C}.\]\[\int\sqrt{1+\frac{x^{2}}{9}}\,dx=\frac{x}{6}\sqrt{x^{2}+9}+\frac{3}{2}\log\left|x+\sqrt{x^{2}+9}\right|+\mathrm{C}\]Equivalently $\displaystyle \dfrac{x}{2}\sqrt{1+\dfrac{x^{2}}{9}}+\dfrac{3}{2}\log\left(x+\sqrt{x^{2}+9}\right)+\mathrm{C}$; it is valid for all real $\displaystyle x$, and the modulus may be dropped because $\displaystyle x+\sqrt{x^{2}+9}>0$ always.
Choose the correct answer in Exercises $\displaystyle 10$ to 11.
Exercise 10
$\displaystyle \int \sqrt{1+x^{2}} d x$ is equal to (A) $\displaystyle \frac{x}{2} \sqrt{1+x^{2}}+\frac{1}{2} \log \left|\left(x+\sqrt{1+x^{2}}\right)\right|+\mathrm{C}$ (B) $\displaystyle \frac{2}{3}\left(1+x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$ (C) $\displaystyle \frac{2}{3} x\left(1+x^{2}\right)^{\frac{3}{2}}+\mathrm{C}$ (D) $\displaystyle \frac{x^{2}}{2} \sqrt{1+x^{2}}+\frac{1}{2} x^{2} \log \left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}$
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NCERT’s answer
A
Use $\displaystyle \int\sqrt{x^{2}+a^{2}}\,dx=\dfrac{x}{2}\sqrt{x^{2}+a^{2}}+\dfrac{a^{2}}{2}\log\left|x+\sqrt{x^{2}+a^{2}}\right|+\mathrm{C}$ with $\displaystyle a=1$: \[\int\sqrt{1+x^{2}}\,dx=\frac{x}{2}\sqrt{1+x^{2}}+\frac{1}{2}\log\left|x+\sqrt{1+x^{2}}\right|+\mathrm{C}.\]The trap here is option (B): $\displaystyle \dfrac{2}{3}\left(1+x^{2}\right)^{3/2}$ is what you would write if the chain rule were ignored, but \[\frac{d}{dx}\left[\frac{2}{3}\left(1+x^{2}\right)^{3/2}\right]=2x\sqrt{1+x^{2}}\neq\sqrt{1+x^{2}},\] because $\displaystyle \sqrt{1+x^{2}}$ is not of the form $\displaystyle f'(x)\left[f(x)\right]^{n}$. Options (C) and (D) fail the same differentiation test.The correct answer is (A).
Exercise 11
$\displaystyle \int \sqrt{x^{2}-8 x+7} d x$ is equal to (A) $\displaystyle \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}+9 \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+\mathrm{C}$ (B) $\displaystyle \frac{1}{2}(x+4) \sqrt{x^{2}-8 x+7}+9 \log \left|x+4+\sqrt{x^{2}-8 x+7}\right|+\mathrm{C}$ (C) $\displaystyle \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-3 \sqrt{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+\mathrm{C}$ (D) $\displaystyle \frac{1}{2}(x-4) \sqrt{x^{2}-8 x+7}-\frac{9}{2} \log \left|x-4+\sqrt{x^{2}-8 x+7}\right|+\mathrm{C}$
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NCERT’s answer
D
Complete the square: \[x^{2}-8x+7=\left(x^{2}-8x+16\right)-9=(x-4)^{2}-3^{2}.\]With $\displaystyle t=x-4,\ dt=dx,\ a=3$, apply $\displaystyle \int\sqrt{t^{2}-a^{2}}\,dt=\dfrac{t}{2}\sqrt{t^{2}-a^{2}}-\dfrac{a^{2}}{2}\log\left|t+\sqrt{t^{2}-a^{2}}\right|+\mathrm{C}$, so that $\displaystyle \dfrac{a^{2}}{2}=\dfrac{9}{2}$ and the log term carries a minus sign: \[\int\sqrt{x^{2}-8x+7}\,dx=\frac{1}{2}(x-4)\sqrt{x^{2}-8x+7}-\frac{9}{2}\log\left|x-4+\sqrt{x^{2}-8x+7}\right|+\mathrm{C}.\]Option (A) has the wrong sign and coefficient on the log, (B) completes the square as $\displaystyle (x+4)^{2}$, and (C) uses $\displaystyle 3\sqrt{2}$ instead of $\displaystyle \dfrac{9}{2}$.The correct answer is (D).

EXERCISE 7.8 1–10 (part 19 of 27)

Evaluate the definite integrals in Exercises $\displaystyle 1$ to 20.
Exercise 1
$\displaystyle \int_{-1}^{1}(x+1) d x$
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NCERT’s answer
$\displaystyle 2$
By the Second Fundamental Theorem of Calculus, if $\displaystyle F$ is an antiderivative of a continuous $\displaystyle f$ on $\displaystyle [a,b]$, then $\displaystyle \int_{a}^{b} f(x)\,dx=F(b)-F(a)$.Here $\displaystyle f(x)=x+1$ is a polynomial, hence continuous, and by the power rule \[F(x)=\frac{x^{2}}{2}+x.\]Substitute the limits. The step to be careful with is the lower one: squaring makes the first term positive while the second stays negative, \[F(1)=\frac{1}{2}+1=\frac{3}{2},\qquad F(-1)=\frac{(-1)^{2}}{2}+(-1)=\frac{1}{2}-1=-\frac{1}{2}.\]Hence \[\int_{-1}^{1}(x+1)\,dx=\frac{3}{2}-\left(-\frac{1}{2}\right)=2.\]The value of the integral is $\displaystyle 2$.
Exercise 2
$\displaystyle \int_{2}^{3} \frac{1}{x} d x$
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NCERT’s answer
$\displaystyle \log \frac{3}{2}$
The antiderivative of $\displaystyle \frac{1}{x}$ is $\displaystyle \log|x|$. On the interval of integration $\displaystyle [2,3]$ we have $\displaystyle x>0$, so $\displaystyle |x|=x$ and the modulus may be dropped; the integrand is also continuous there, so the Second Fundamental Theorem of Calculus applies.\[\int_{2}^{3}\frac{1}{x}\,dx=\bigl[\log x\bigr]_{2}^{3}=\log 3-\log 2.\]Using $\displaystyle \log a-\log b=\log\dfrac{a}{b}$, \[\int_{2}^{3}\frac{1}{x}\,dx=\log\frac{3}{2}.\]
Exercise 3
$\displaystyle \int_{1}^{2}\left(4 x^{3}-5 x^{2}+6 x+9\right) d x$
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NCERT’s answer
$\displaystyle \frac{64}{3}$
Integrate term by term using the power rule $\displaystyle \int x^{n}\,dx=\dfrac{x^{n+1}}{n+1}\ (n\neq-1)$, then apply the Second Fundamental Theorem of Calculus.An antiderivative is \[F(x)=x^{4}-\frac{5x^{3}}{3}+3x^{2}+9x.\]Evaluate at the limits: \[F(2)=16-\frac{40}{3}+12+18=46-\frac{40}{3},\] \[F(1)=1-\frac{5}{3}+3+9=13-\frac{5}{3}.\]Subtracting, and keeping the two fractional parts together so the arithmetic stays clean, \[\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)dx=(46-13)-\frac{40-5}{3}=33-\frac{35}{3}=\frac{99-35}{3}.\]\[\int_{1}^{2}\left(4x^{3}-5x^{2}+6x+9\right)dx=\frac{64}{3}\]
Exercise 4
$\displaystyle \int_{0}^{\frac{\pi}{4}} \sin 2 x d x$
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NCERT’s answer
$\displaystyle \frac{1}{2}$
By the chain rule, $\displaystyle \dfrac{d}{dx}\left(-\dfrac{\cos 2x}{2}\right)=\sin 2x$, so \[\int \sin 2x\,dx=-\frac{\cos 2x}{2}.\] The factor $\displaystyle \tfrac12$ coming from the inner derivative $\displaystyle 2$ is the step most often dropped.By the Second Fundamental Theorem of Calculus, \[\int_{0}^{\pi/4}\sin 2x\,dx=\left[-\frac{\cos 2x}{2}\right]_{0}^{\pi/4}=-\frac{\cos\dfrac{\pi}{2}}{2}+\frac{\cos 0}{2}.\]Since $\displaystyle \cos\frac{\pi}{2}=0$ and $\displaystyle \cos 0=1$, \[\int_{0}^{\pi/4}\sin 2x\,dx=0+\frac{1}{2}=\frac{1}{2}.\]
Exercise 5
$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos 2 x d x$
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NCERT’s answer
$\displaystyle 0$
Since $\displaystyle \dfrac{d}{dx}\left(\dfrac{\sin 2x}{2}\right)=\cos 2x$, \[\int \cos 2x\,dx=\frac{\sin 2x}{2}.\]Applying the Second Fundamental Theorem of Calculus, and remembering that the upper limit $\displaystyle x=\frac{\pi}{2}$ must be doubled inside the sine (it gives $\displaystyle \sin\pi$, not $\displaystyle \sin\frac{\pi}{2}$), \[\int_{0}^{\pi/2}\cos 2x\,dx=\left[\frac{\sin 2x}{2}\right]_{0}^{\pi/2}=\frac{\sin \pi}{2}-\frac{\sin 0}{2}=\frac{0}{2}-\frac{0}{2}.\]\[\int_{0}^{\pi/2}\cos 2x\,dx=0\]
Exercise 6
$\displaystyle \int_{4}^{5} e^{x} d x$
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NCERT’s answer
$\displaystyle e^{4}(e-1)$
The exponential function is its own antiderivative: $\displaystyle \int e^{x}\,dx=e^{x}$. By the Second Fundamental Theorem of Calculus, \[\int_{4}^{5}e^{x}\,dx=\bigl[e^{x}\bigr]_{4}^{5}=e^{5}-e^{4}.\]Taking out the common factor $\displaystyle e^{4}$, \[\int_{4}^{5}e^{x}\,dx=e^{4}(e-1).\]
Exercise 7
$\displaystyle \int_{0}^{\frac{\pi}{4}} \tan x d x$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log 2$
The standard result is $\displaystyle \int \tan x\,dx=\log|\sec x|$ (obtained by putting $\displaystyle t=\cos x$). On $\displaystyle \left[0,\frac{\pi}{4}\right]$ we have $\displaystyle \cos x>0$, so $\displaystyle \sec x>0$ and the modulus can be dropped; the integrand is continuous there, so the Second Fundamental Theorem of Calculus applies.\[\int_{0}^{\pi/4}\tan x\,dx=\bigl[\log \sec x\bigr]_{0}^{\pi/4}=\log\sec\frac{\pi}{4}-\log\sec 0.\]Now $\displaystyle \sec\frac{\pi}{4}=\sqrt{2}$ and $\displaystyle \sec 0=1$, so \[\int_{0}^{\pi/4}\tan x\,dx=\log\sqrt{2}-\log 1=\frac{1}{2}\log 2.\]
Exercise 8
$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{4}} \operatorname{cosec} x d x$
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NCERT’s answer
$\displaystyle \log \left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)$
The standard result is \[\int \mathrm{cosec}\,x\,dx=\log\left|\mathrm{cosec}\,x-\cot x\right|.\] On $\displaystyle \left[\frac{\pi}{6},\frac{\pi}{4}\right]$ the integrand is continuous, so the Second Fundamental Theorem of Calculus applies.At the upper limit $\displaystyle x=\frac{\pi}{4}$: $\displaystyle \mathrm{cosec}\,\frac{\pi}{4}=\sqrt{2}$, $\displaystyle \cot\frac{\pi}{4}=1$, so the bracket is $\displaystyle \sqrt{2}-1>0$.At the lower limit $\displaystyle x=\frac{\pi}{6}$: $\displaystyle \mathrm{cosec}\,\frac{\pi}{6}=2$, $\displaystyle \cot\frac{\pi}{6}=\sqrt{3}$, so the bracket is $\displaystyle 2-\sqrt{3}>0$.Both are positive, so the moduli may be dropped: \[\int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx=\log(\sqrt{2}-1)-\log(2-\sqrt{3}).\]\[\int_{\pi/6}^{\pi/4}\mathrm{cosec}\,x\,dx=\log\left(\frac{\sqrt{2}-1}{2-\sqrt{3}}\right)\](Equivalently $\displaystyle \log\dfrac{\tan\frac{\pi}{8}}{\tan\frac{\pi}{12}}$, since $\displaystyle \mathrm{cosec}\,x-\cot x=\tan\frac{x}{2}$.)
Exercise 9
$\displaystyle \int_{0}^{1} \frac{d x}{\sqrt{1-x^{2}}}$
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NCERT’s answer
$\displaystyle \frac{\pi}{2}$
By definition of the inverse sine, $\displaystyle \dfrac{d}{dx}\sin^{-1}x=\dfrac{1}{\sqrt{1-x^{2}}}$ for $\displaystyle x\in(-1,1)$, so $\displaystyle \sin^{-1}x$ is an antiderivative of the integrand.The integrand itself is unbounded as $\displaystyle x\to 1^{-}$, but $\displaystyle \sin^{-1}x$ remains continuous up to $\displaystyle x=1$, so the value of the integral is obtained by substituting the limits into $\displaystyle \sin^{-1}x$ as usual.Using principal values, $\displaystyle \sin^{-1}1=\frac{\pi}{2}$ (the value in $\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$) and $\displaystyle \sin^{-1}0=0$: \[\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\bigl[\sin^{-1}x\bigr]_{0}^{1}=\frac{\pi}{2}-0.\]\[\int_{0}^{1}\frac{dx}{\sqrt{1-x^{2}}}=\frac{\pi}{2}\]
Exercise 10
$\displaystyle \int_{0}^{1} \frac{d x}{1+x^{2}}$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
Since $\displaystyle \dfrac{d}{dx}\tan^{-1}x=\dfrac{1}{1+x^{2}}$, the function $\displaystyle \tan^{-1}x$ is an antiderivative of the integrand, which is continuous on $\displaystyle [0,1]$. By the Second Fundamental Theorem of Calculus, \[\int_{0}^{1}\frac{dx}{1+x^{2}}=\bigl[\tan^{-1}x\bigr]_{0}^{1}=\tan^{-1}1-\tan^{-1}0.\]Taking principal values (branch $\displaystyle \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$), $\displaystyle \tan^{-1}1=\frac{\pi}{4}$ and $\displaystyle \tan^{-1}0=0$: \[\int_{0}^{1}\frac{dx}{1+x^{2}}=\frac{\pi}{4}.\]

EXERCISE 7.8 11–22 (part 20 of 27)

Evaluate the definite integrals in Exercises $\displaystyle 1$ to 20.
Exercise 11
$\displaystyle \int_{2}^{3} \frac{d x}{x^{2}-1}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \frac{3}{2}$
Factorise the denominator and split by partial fractions: \[\frac{1}{x^{2}-1}=\frac{1}{(x-1)(x+1)}=\frac{A}{x-1}+\frac{B}{x+1}.\] Then $\displaystyle 1=A(x+1)+B(x-1)$; putting $\displaystyle x=1$ gives $\displaystyle A=\frac{1}{2}$ and $\displaystyle x=-1$ gives $\displaystyle B=-\frac{1}{2}$. So \[\frac{1}{x^{2}-1}=\frac{1}{2}\left(\frac{1}{x-1}-\frac{1}{x+1}\right).\]On $\displaystyle [2,3]$ both $\displaystyle x-1$ and $\displaystyle x+1$ are positive, so the moduli drop, and the integrand is continuous there: \[\int_{2}^{3}\frac{dx}{x^{2}-1}=\frac{1}{2}\left[\log\frac{x-1}{x+1}\right]_{2}^{3}=\frac{1}{2}\left(\log\frac{2}{4}-\log\frac{1}{3}\right).\]Since $\displaystyle \log\frac{1}{2}-\log\frac{1}{3}=\log\frac{3}{2}$, \[\int_{2}^{3}\frac{dx}{x^{2}-1}=\frac{1}{2}\log\frac{3}{2}.\]
Exercise 12
$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
A power of cosine cannot be integrated directly; use the double-angle identity $\displaystyle \cos 2x=2\cos^{2}x-1$, i.e. \[\cos^{2}x=\frac{1+\cos 2x}{2}.\]Hence \[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\int_{0}^{\pi/2}(1+\cos 2x)\,dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]At $\displaystyle x=\frac{\pi}{2}$ the sine term is $\displaystyle \frac{\sin\pi}{2}=0$, and at $\displaystyle x=0$ everything vanishes, so \[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\left(\frac{\pi}{2}+0\right)=\frac{\pi}{4}.\]
Exercise 13
$\displaystyle \int_{2}^{3} \frac{x d x}{x^{2}+1}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log 2$
The numerator is (a constant times) the derivative of the denominator, so use the substitution \[t=x^{2}+1,\qquad dt=2x\,dx\ \Rightarrow\ x\,dx=\frac{dt}{2}.\]Change the limits with the variable (this is the step usually forgotten): $\displaystyle x=2\Rightarrow t=5$ and $\displaystyle x=3\Rightarrow t=10$. Then \[\int_{2}^{3}\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\int_{5}^{10}\frac{dt}{t}=\frac{1}{2}\bigl[\log t\bigr]_{5}^{10}=\frac{1}{2}\left(\log 10-\log 5\right).\]\[\int_{2}^{3}\frac{x\,dx}{x^{2}+1}=\frac{1}{2}\log 2\]
Exercise 14
$\displaystyle \int_{0}^{1} \frac{2 x+3}{5 x^{2}+1} d x$
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NCERT’s answer
$\displaystyle \frac{1}{5} \log 6+\frac{3}{\sqrt{5}} \tan ^{-1} \sqrt{5}$
Split the numerator into the part that is proportional to the derivative of the denominator and a constant part: \[\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\,dx=\int_{0}^{1}\frac{2x}{5x^{2}+1}\,dx+3\int_{0}^{1}\frac{dx}{5x^{2}+1}=I_{1}+I_{2}.\]$\displaystyle I_{1}$: put $\displaystyle t=5x^{2}+1$, so $\displaystyle dt=10x\,dx$, i.e. $\displaystyle 2x\,dx=\frac{dt}{5}$. The limits become $\displaystyle x=0\Rightarrow t=1$, $\displaystyle x=1\Rightarrow t=6$: \[I_{1}=\frac{1}{5}\int_{1}^{6}\frac{dt}{t}=\frac{1}{5}\bigl[\log t\bigr]_{1}^{6}=\frac{1}{5}\log 6.\]$\displaystyle I_{2}$: take the coefficient of $\displaystyle x^{2}$ out first, then use $\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}$ with $\displaystyle a=\frac{1}{\sqrt5}$: \[I_{2}=\frac{3}{5}\int_{0}^{1}\frac{dx}{x^{2}+\left(\dfrac{1}{\sqrt5}\right)^{2}}=\frac{3}{5}\cdot\sqrt5\left[\tan^{-1}\left(\sqrt5\,x\right)\right]_{0}^{1}=\frac{3}{\sqrt5}\tan^{-1}\sqrt5 .\]Adding, \[\int_{0}^{1}\frac{2x+3}{5x^{2}+1}\,dx=\frac{1}{5}\log 6+\frac{3}{\sqrt{5}}\tan^{-1}\sqrt{5}.\]
Exercise 15
$\displaystyle \int_{0}^{1} x e^{x^{2}} d x$
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NCERT’s answer
$\displaystyle \frac{1}{2}(e-1)$
The factor $\displaystyle x$ outside is, up to a constant, the derivative of the exponent $\displaystyle x^{2}$, so substitute \[t=x^{2},\qquad dt=2x\,dx\ \Rightarrow\ x\,dx=\frac{dt}{2}.\]Change the limits along with the variable: $\displaystyle x=0\Rightarrow t=0$ and $\displaystyle x=1\Rightarrow t=1$. Then \[\int_{0}^{1}x\,e^{x^{2}}dx=\frac{1}{2}\int_{0}^{1}e^{t}\,dt=\frac{1}{2}\bigl[e^{t}\bigr]_{0}^{1}=\frac{1}{2}\left(e^{1}-e^{0}\right).\]\[\int_{0}^{1}x\,e^{x^{2}}dx=\frac{e-1}{2}\]
Exercise 16
$\displaystyle \int_{1}^{2} \frac{5 x^{2}}{x^{2}+4 x+3}$
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NCERT’s answer
$\displaystyle 5-\frac{5}{2}\left(9 \log \frac{5}{4}-\log \frac{3}{2}\right)$
The integrand $\displaystyle \dfrac{5x^{2}}{x^{2}+4x+3}$ is an improper rational function (degree of numerator = degree of denominator), so divide first: \[5x^{2}=5\left(x^{2}+4x+3\right)-(20x+15)\ \Longrightarrow\ \frac{5x^{2}}{x^{2}+4x+3}=5-\frac{20x+15}{(x+1)(x+3)}.\]Now split the proper part by partial fractions: \[\frac{20x+15}{(x+1)(x+3)}=\frac{A}{x+1}+\frac{B}{x+3},\qquad 20x+15=A(x+3)+B(x+1).\] Putting $\displaystyle x=-1$: $\displaystyle -5=2A\Rightarrow A=-\frac{5}{2}$. Putting $\displaystyle x=-3$: $\displaystyle -45=-2B\Rightarrow B=\frac{45}{2}$.So the integrand becomes \[5+\frac{5}{2}\cdot\frac{1}{x+1}-\frac{45}{2}\cdot\frac{1}{x+3}\] (note the two sign changes: the partial fractions are subtracted from $\displaystyle 5$).On $\displaystyle [1,2]$ both $\displaystyle x+1$ and $\displaystyle x+3$ are positive, so the moduli drop: \[\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx=\left[5x+\frac{5}{2}\log (x+1)-\frac{45}{2}\log (x+3)\right]_{1}^{2}\] \[=\left(10+\frac{5}{2}\log 3-\frac{45}{2}\log 5\right)-\left(5+\frac{5}{2}\log 2-\frac{45}{2}\log 4\right).\]Collecting the logarithms, \[\int_{1}^{2}\frac{5x^{2}}{x^{2}+4x+3}\,dx=5+\frac{5}{2}\log\frac{3}{2}-\frac{45}{2}\log\frac{5}{4}.\]
Exercise 17
$\displaystyle \int_{0}^{\frac{\pi}{4}}\left(2 \sec ^{2} x+x^{3}+2\right) d x$
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NCERT’s answer
$\displaystyle \frac{\pi^{4}}{1024}+\frac{\pi}{2}+2$
Integrate term by term, using $\displaystyle \int \sec^{2}x\,dx=\tan x$ and the power rule.An antiderivative is \[F(x)=2\tan x+\frac{x^{4}}{4}+2x.\]By the Second Fundamental Theorem of Calculus, with $\displaystyle F(0)=0$ and $\displaystyle \tan\frac{\pi}{4}=1$, \[\int_{0}^{\pi/4}\left(2\sec^{2}x+x^{3}+2\right)dx=2(1)+\frac{1}{4}\left(\frac{\pi}{4}\right)^{4}+2\cdot\frac{\pi}{4}.\]Since $\displaystyle \dfrac{1}{4}\cdot\dfrac{\pi^{4}}{256}=\dfrac{\pi^{4}}{1024}$, \[\int_{0}^{\pi/4}\left(2\sec^{2}x+x^{3}+2\right)dx=2+\frac{\pi}{2}+\frac{\pi^{4}}{1024}.\]
Exercise 18
$\displaystyle \int_{0}^{\pi}\left(\sin ^{2} \frac{x}{2}-\cos ^{2} \frac{x}{2}\right) d x$
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NCERT’s answer
$\displaystyle 0$
Do not integrate the two squares separately; recognise the double-angle identity \[\cos\theta=\cos^{2}\frac{\theta}{2}-\sin^{2}\frac{\theta}{2}\ \Longrightarrow\ \sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}=-\cos x.\]Hence \[\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx=-\int_{0}^{\pi}\cos x\,dx=-\bigl[\sin x\bigr]_{0}^{\pi}=-\left(\sin\pi-\sin 0\right).\]Both sines vanish, so \[\int_{0}^{\pi}\left(\sin^{2}\frac{x}{2}-\cos^{2}\frac{x}{2}\right)dx=0.\]
Exercise 19
$\displaystyle \int_{0}^{2} \frac{6 x+3}{x^{2}+4} d x$
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NCERT’s answer
$\displaystyle 3 \log 2+\frac{3 \pi}{8}$
Split the numerator so that one part is a multiple of $\displaystyle \frac{d}{dx}\left(x^{2}+4\right)=2x$: \[\int_{0}^{2}\frac{6x+3}{x^{2}+4}\,dx=3\int_{0}^{2}\frac{2x}{x^{2}+4}\,dx+3\int_{0}^{2}\frac{dx}{x^{2}+4}.\]For the first integral put $\displaystyle t=x^{2}+4$ (so $\displaystyle dt=2x\,dx$, limits $\displaystyle t=4$ to $\displaystyle t=8$): \[3\int_{4}^{8}\frac{dt}{t}=3\bigl[\log t\bigr]_{4}^{8}=3\left(\log 8-\log 4\right)=3\log 2.\]For the second use $\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{x}{a}$ with $\displaystyle a=2$: \[3\cdot\frac{1}{2}\left[\tan^{-1}\frac{x}{2}\right]_{0}^{2}=\frac{3}{2}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{3}{2}\cdot\frac{\pi}{4}=\frac{3\pi}{8}.\]Adding, \[\int_{0}^{2}\frac{6x+3}{x^{2}+4}\,dx=3\log 2+\frac{3\pi}{8}.\]
Exercise 20
$\displaystyle \int_{0}^{1}\left(x e^{x}+\sin \frac{\pi x}{4}\right) d x$
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NCERT’s answer
$\displaystyle 1+\frac{4}{\pi}-\frac{2 \sqrt{2}}{\pi}$
Split the integral into two parts.Part $\displaystyle 1$: $\displaystyle \displaystyle\int_{0}^{1}x\,e^{x}dx$ needs integration by parts, $\displaystyle \int u\,v'\,dx=uv-\int u'v\,dx$. By ILATE take $\displaystyle u=x$ (algebraic) and $\displaystyle v'=e^{x}$: \[\int x\,e^{x}dx=x e^{x}-\int e^{x}dx=e^{x}(x-1).\] Hence \[\int_{0}^{1}x\,e^{x}dx=\bigl[e^{x}(x-1)\bigr]_{0}^{1}=e^{1}(0)-e^{0}(-1)=1.\]Part $\displaystyle 2$: $\displaystyle \displaystyle\int_{0}^{1}\sin\frac{\pi x}{4}\,dx$; the chain-rule factor is $\displaystyle \frac{4}{\pi}$: \[\int_{0}^{1}\sin\frac{\pi x}{4}\,dx=\left[-\frac{4}{\pi}\cos\frac{\pi x}{4}\right]_{0}^{1}=-\frac{4}{\pi}\left(\cos\frac{\pi}{4}-\cos 0\right)=-\frac{4}{\pi}\left(\frac{1}{\sqrt{2}}-1\right).\] That is $\displaystyle \dfrac{4}{\pi}-\dfrac{4}{\pi\sqrt{2}}=\dfrac{4}{\pi}-\dfrac{2\sqrt{2}}{\pi}$.Adding the two parts, \[\int_{0}^{1}\left(x e^{x}+\sin\frac{\pi x}{4}\right)dx=1+\frac{4}{\pi}-\frac{2\sqrt{2}}{\pi}.\]
Choose the correct answer in Exercises $\displaystyle 21$ and 22.
Exercise 21
$\displaystyle \int_{1}^{\sqrt{3}} \frac{d x}{1+x^{2}}$ equals (A) $\displaystyle \frac{\pi}{3}$ (B) $\displaystyle \frac{2 \pi}{3}$ (C) $\displaystyle \frac{\pi}{6}$ (D) $\displaystyle \frac{\pi}{12}$
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NCERT’s answer
D
The antiderivative of $\displaystyle \dfrac{1}{1+x^{2}}$ is $\displaystyle \tan^{-1}x$, and the integrand is continuous on $\displaystyle \left[1,\sqrt{3}\right]$, so by the Second Fundamental Theorem of Calculus \[\int_{1}^{\sqrt{3}}\frac{dx}{1+x^{2}}=\bigl[\tan^{-1}x\bigr]_{1}^{\sqrt{3}}=\tan^{-1}\sqrt{3}-\tan^{-1}1.\]Using principal values, $\displaystyle \tan^{-1}\sqrt{3}=\frac{\pi}{3}$ and $\displaystyle \tan^{-1}1=\frac{\pi}{4}$. The trap here is stopping at $\displaystyle \frac{\pi}{3}$ and forgetting the lower limit: \[\frac{\pi}{3}-\frac{\pi}{4}=\frac{4\pi-3\pi}{12}=\frac{\pi}{12}.\]The value is $\displaystyle \dfrac{\pi}{12}$, so the correct option is (D).
Exercise 22
$\displaystyle \int_{0}^{\frac{2}{3}} \frac{d x}{4+9 x^{2}}$ equals (A) $\displaystyle \frac{\pi}{6}$ (B) $\displaystyle \frac{\pi}{12}$ (C) $\displaystyle \frac{\pi}{24}$ (D) $\displaystyle \frac{\pi}{4}$
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NCERT’s answer
C
Take the coefficient of $\displaystyle x^{2}$ outside so the standard form $\displaystyle \int\dfrac{dx}{x^{2}+a^{2}}=\dfrac{1}{a}\tan^{-1}\dfrac{x}{a}$ can be used: \[\frac{1}{4+9x^{2}}=\frac{1}{9}\cdot\frac{1}{x^{2}+\left(\dfrac{2}{3}\right)^{2}}.\]With $\displaystyle a=\frac{2}{3}$, \[\int\frac{dx}{4+9x^{2}}=\frac{1}{9}\cdot\frac{3}{2}\tan^{-1}\frac{3x}{2}=\frac{1}{6}\tan^{-1}\frac{3x}{2}.\]Now apply the limits $\displaystyle x=0$ to $\displaystyle x=\frac{2}{3}$; at the upper limit $\displaystyle \frac{3x}{2}=1$: \[\int_{0}^{2/3}\frac{dx}{4+9x^{2}}=\frac{1}{6}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{1}{6}\cdot\frac{\pi}{4}=\frac{\pi}{24}.\]The value is $\displaystyle \dfrac{\pi}{24}$, so the correct option is (C).

EXERCISE 7.9 1–10 (part 21 of 27)

Evaluate the integrals in Exercises $\displaystyle 1$ to $\displaystyle 8$ using substitution.
Exercise 1
$\displaystyle \int_{0}^{1} \frac{x}{x^{2}+1} d x$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log 2$
Substitution in a definite integral. Put $\displaystyle t=x^{2}+1$, so that $\displaystyle dt=2x\,dx$, i.e. $\displaystyle x\,dx=\tfrac{1}{2}\,dt$. Change the limits along with the variable — $\displaystyle x=0\Rightarrow t=1$ and $\displaystyle x=1\Rightarrow t=2$ — so that no back-substitution is needed. \[\int_{0}^{1}\frac{x}{x^{2}+1}\,dx=\frac{1}{2}\int_{1}^{2}\frac{dt}{t}=\frac{1}{2}\Big[\log|t|\Big]_{1}^{2}=\frac{1}{2}\left(\log 2-\log 1\right)\] Final answer: $\displaystyle \dfrac{1}{2}\log 2$.
Exercise 2
$\displaystyle \int_{0}^{\frac{\pi}{2}} \sqrt{\sin \phi} \cos ^{5} \phi d \phi$
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NCERT’s answer
$\displaystyle \frac{64}{231}$
The odd power of $\displaystyle \cos\phi$ is the cue: keep one factor $\displaystyle \cos\phi\,d\phi$ for the substitution and convert the rest using $\displaystyle \cos^{2}\phi=1-\sin^{2}\phi$. Put $\displaystyle \sin\phi=t$, so $\displaystyle \cos\phi\,d\phi=dt$. Limits: $\displaystyle \phi=0\Rightarrow t=0$ and $\displaystyle \phi=\frac{\pi}{2}\Rightarrow t=1$. \[\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\cos^{5}\phi\,d\phi=\int_{0}^{\frac{\pi}{2}}\sqrt{\sin\phi}\,\left(1-\sin^{2}\phi\right)^{2}\cos\phi\,d\phi=\int_{0}^{1}t^{\frac{1}{2}}\left(1-t^{2}\right)^{2}dt\] Expand before integrating: \[\int_{0}^{1}\left(t^{\frac{1}{2}}-2t^{\frac{5}{2}}+t^{\frac{9}{2}}\right)dt=\left[\frac{2}{3}t^{\frac{3}{2}}-\frac{4}{7}t^{\frac{7}{2}}+\frac{2}{11}t^{\frac{11}{2}}\right]_{0}^{1}=\frac{2}{3}-\frac{4}{7}+\frac{2}{11}\] \[=\frac{154-132+42}{231}=\frac{64}{231}\] Final answer: $\displaystyle \dfrac{64}{231}$.
Exercise 3
$\displaystyle \int_{0}^{1} \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{2}-\log 2$
Simplify the inverse-trigonometric function first. Put $\displaystyle x=\tan\theta$, i.e. $\displaystyle \theta=\tan^{-1}x$; then \[\frac{2x}{1+x^{2}}=\frac{2\tan\theta}{1+\tan^{2}\theta}=\sin 2\theta\] The domain check is the step that decides the answer. As $\displaystyle x$ runs over $\displaystyle [0,1]$, $\displaystyle \theta=\tan^{-1}x$ runs over $\displaystyle \left[0,\frac{\pi}{4}\right]$, so $\displaystyle 2\theta\in\left[0,\frac{\pi}{2}\right]$, which lies inside the principal range $\displaystyle \left[-\frac{\pi}{2},\frac{\pi}{2}\right]$ of $\displaystyle \sin^{-1}$. Only for that reason may we write $\displaystyle \sin^{-1}(\sin 2\theta)=2\theta$. Hence on $\displaystyle [0,1]$ \[\sin^{-1}\left(\frac{2x}{1+x^{2}}\right)=2\tan^{-1}x\] So $\displaystyle \mathrm{I}=2\displaystyle\int_{0}^{1}\tan^{-1}x\,dx$. Integrate by parts taking $\displaystyle \tan^{-1}x$ as the first function and $\displaystyle 1$ as the second: \[\int\tan^{-1}x\,dx=x\tan^{-1}x-\int\frac{x}{1+x^{2}}dx=x\tan^{-1}x-\frac{1}{2}\log\left(1+x^{2}\right)\] Therefore \[\mathrm{I}=2\left[x\tan^{-1}x-\frac{1}{2}\log\left(1+x^{2}\right)\right]_{0}^{1}=2\left[\left(\frac{\pi}{4}-\frac{1}{2}\log 2\right)-\left(0-0\right)\right]\] Final answer: $\displaystyle \dfrac{\pi}{2}-\log 2$.
Exercise 4
$\displaystyle \int_{0}^{2} x \sqrt{x+2}$ (Put $\displaystyle x+2=t^{2}$ )
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NCERT’s answer
$\displaystyle \frac{16 \sqrt{2}}{15}(\sqrt{2}+1)$
Follow the hint. Put $\displaystyle x+2=t^{2}$, so $\displaystyle x=t^{2}-2$, $\displaystyle dx=2t\,dt$, and $\displaystyle \sqrt{x+2}=t$ (the positive root, since $\displaystyle t>0$ over the range). The limits change too: $\displaystyle x=0\Rightarrow t^{2}=2\Rightarrow t=\sqrt{2}$, and $\displaystyle x=2\Rightarrow t^{2}=4\Rightarrow t=2$. \[\int_{0}^{2}x\sqrt{x+2}\,dx=\int_{\sqrt{2}}^{2}\left(t^{2}-2\right)\cdot t\cdot 2t\,dt=2\int_{\sqrt{2}}^{2}\left(t^{4}-2t^{2}\right)dt\] \[=2\left[\frac{t^{5}}{5}-\frac{2t^{3}}{3}\right]_{\sqrt{2}}^{2}\] Use $\displaystyle \left(\sqrt{2}\right)^{5}=4\sqrt{2}$ and $\displaystyle \left(\sqrt{2}\right)^{3}=2\sqrt{2}$: \[=2\left[\left(\frac{32}{5}-\frac{16}{3}\right)-\left(\frac{4\sqrt{2}}{5}-\frac{4\sqrt{2}}{3}\right)\right]=2\left[\frac{96-80}{15}-\left(-\frac{8\sqrt{2}}{15}\right)\right]\] \[=2\left[\frac{16}{15}+\frac{8\sqrt{2}}{15}\right]=\frac{32+16\sqrt{2}}{15}\] Final answer: $\displaystyle \dfrac{16\left(2+\sqrt{2}\right)}{15}$.
Exercise 5
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin x}{1+\cos ^{2} x} d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
The numerator is, up to sign, the derivative of $\displaystyle \cos x$, so substitute $\displaystyle \cos x=t$, giving $\displaystyle -\sin x\,dx=dt$. Limits: $\displaystyle x=0\Rightarrow t=1$ and $\displaystyle x=\frac{\pi}{2}\Rightarrow t=0$. \[\int_{0}^{\frac{\pi}{2}}\frac{\sin x}{1+\cos^{2}x}\,dx=-\int_{1}^{0}\frac{dt}{1+t^{2}}=\int_{0}^{1}\frac{dt}{1+t^{2}}\] (the swap of limits absorbs the minus sign). \[=\Big[\tan^{-1}t\Big]_{0}^{1}=\tan^{-1}1-\tan^{-1}0=\frac{\pi}{4}-0\] Final answer: $\displaystyle \dfrac{\pi}{4}$.
Exercise 6
$\displaystyle \int_{0}^{2} \frac{d x}{x+4-x^{2}}$
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NCERT’s answer
$\displaystyle \frac{1}{\sqrt{17}} \log \frac{21+5 \sqrt{17}}{4}$
Complete the square in the denominator so the integral takes a standard form. \[x+4-x^{2}=-\left(x^{2}-x-4\right)=-\left[\left(x-\frac{1}{2}\right)^{2}-\frac{1}{4}-4\right]=\left(\frac{\sqrt{17}}{2}\right)^{2}-\left(x-\frac{1}{2}\right)^{2}\] This is $\displaystyle a^{2}-u^{2}$ with $\displaystyle a=\frac{\sqrt{17}}{2}$ and $\displaystyle u=x-\frac{1}{2}$, so use \[\int\frac{du}{a^{2}-u^{2}}=\frac{1}{2a}\log\left|\frac{a+u}{a-u}\right|+\mathrm{C},\qquad \frac{1}{2a}=\frac{1}{\sqrt{17}}\] Multiplying numerator and denominator inside the logarithm by $\displaystyle 2$, \[\int_{0}^{2}\frac{dx}{x+4-x^{2}}=\frac{1}{\sqrt{17}}\left[\log\left|\frac{\sqrt{17}+2x-1}{\sqrt{17}-2x+1}\right|\right]_{0}^{2}\] \[=\frac{1}{\sqrt{17}}\left[\log\frac{\sqrt{17}+3}{\sqrt{17}-3}-\log\frac{\sqrt{17}-1}{\sqrt{17}+1}\right]=\frac{1}{\sqrt{17}}\log\frac{\left(\sqrt{17}+3\right)\left(\sqrt{17}+1\right)}{\left(\sqrt{17}-3\right)\left(\sqrt{17}-1\right)}\] \[=\frac{1}{\sqrt{17}}\log\frac{20+4\sqrt{17}}{20-4\sqrt{17}}=\frac{1}{\sqrt{17}}\log\frac{5+\sqrt{17}}{5-\sqrt{17}}\] The quadratic $\displaystyle x+4-x^{2}$ is positive throughout $\displaystyle [0,2]$ (its least value on the interval is $\displaystyle 2$, at $\displaystyle x=2$), so there is no singularity inside the range and the modulus signs may be dropped. Final answer: $\displaystyle \dfrac{1}{\sqrt{17}}\log\dfrac{5+\sqrt{17}}{5-\sqrt{17}}$, which on rationalising equals $\displaystyle \dfrac{1}{\sqrt{17}}\log\dfrac{21+5\sqrt{17}}{4}\approx 0.568$.
Exercise 7
$\displaystyle \int_{-1}^{1} \frac{d x}{x^{2}+2 x+5}$
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NCERT’s answer
$\displaystyle \frac{\pi}{8}$
Complete the square: $\displaystyle x^{2}+2x+5=(x+1)^{2}+2^{2}$. Put $\displaystyle u=x+1$, so $\displaystyle du=dx$; the limits shift to $\displaystyle x=-1\Rightarrow u=0$ and $\displaystyle x=1\Rightarrow u=2$. \[\int_{-1}^{1}\frac{dx}{x^{2}+2x+5}=\int_{0}^{2}\frac{du}{u^{2}+2^{2}}\] By the standard result $\displaystyle \int\frac{du}{u^{2}+a^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+\mathrm{C}$ with $\displaystyle a=2$, \[=\frac{1}{2}\left[\tan^{-1}\frac{u}{2}\right]_{0}^{2}=\frac{1}{2}\left(\tan^{-1}1-\tan^{-1}0\right)=\frac{1}{2}\cdot\frac{\pi}{4}\] Final answer: $\displaystyle \dfrac{\pi}{8}$.
Exercise 8
$\displaystyle \int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2 x^{2}}\right) e^{2 x} d x$
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NCERT’s answer
$\displaystyle \frac{e^{2}\left(e^{2}-2\right)}{4}$
Recognise the standard form $\displaystyle \displaystyle\int e^{ax}\left[a\,f(x)+f'(x)\right]dx=e^{ax}f(x)+\mathrm{C}$. Here $\displaystyle a=2$. Try $\displaystyle f(x)=\dfrac{1}{2x}$, so that $\displaystyle f'(x)=-\dfrac{1}{2x^{2}}$, and check: \[a\,f(x)+f'(x)=2\cdot\frac{1}{2x}-\frac{1}{2x^{2}}=\frac{1}{x}-\frac{1}{2x^{2}}\] which is exactly the bracket in the question. Hence an antiderivative is $\displaystyle \dfrac{e^{2x}}{2x}$, and by the second fundamental theorem of calculus (the integrand is continuous on $\displaystyle [1,2]$, since $\displaystyle x\neq 0$ there), \[\int_{1}^{2}\left(\frac{1}{x}-\frac{1}{2x^{2}}\right)e^{2x}dx=\left[\frac{e^{2x}}{2x}\right]_{1}^{2}=\frac{e^{4}}{4}-\frac{e^{2}}{2}\] Final answer: $\displaystyle \dfrac{e^{4}}{4}-\dfrac{e^{2}}{2}=\dfrac{e^{2}}{4}\left(e^{2}-2\right)$.
Choose the correct answer in Exercises $\displaystyle 9$ and 10.
Exercise 9
The value of the integral $\displaystyle \int_{\frac{1}{3}}^{1} \frac{\left(x-x^{3}\right)^{\frac{1}{3}}}{x^{4}} d x$ is (A) $\displaystyle 6$ (B) $\displaystyle 0$ (C) $\displaystyle 3$ (D) $\displaystyle 4$
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NCERT’s answer
D
On $\displaystyle \left[\frac{1}{3},1\right]$ we have $\displaystyle x>0$, so take $\displaystyle x^{3}$ — not $\displaystyle x$ — out of the cube root; that is the step that makes the substitution work. \[\left(x-x^{3}\right)^{\frac{1}{3}}=\left[x^{3}\left(\frac{1}{x^{2}}-1\right)\right]^{\frac{1}{3}}=x\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}\] Therefore \[\mathrm{I}=\int_{\frac{1}{3}}^{1}\frac{x\left(\dfrac{1}{x^{2}}-1\right)^{\frac{1}{3}}}{x^{4}}dx=\int_{\frac{1}{3}}^{1}\left(\frac{1}{x^{2}}-1\right)^{\frac{1}{3}}\frac{dx}{x^{3}}\] Put $\displaystyle t=\dfrac{1}{x^{2}}-1$; then $\displaystyle dt=-\dfrac{2}{x^{3}}dx$, i.e. $\displaystyle \dfrac{dx}{x^{3}}=-\dfrac{dt}{2}$. Limits: $\displaystyle x=\frac{1}{3}\Rightarrow t=9-1=8$, and $\displaystyle x=1\Rightarrow t=0$. \[\mathrm{I}=-\frac{1}{2}\int_{8}^{0}t^{\frac{1}{3}}dt=\frac{1}{2}\int_{0}^{8}t^{\frac{1}{3}}dt=\frac{1}{2}\cdot\frac{3}{4}\left[t^{\frac{4}{3}}\right]_{0}^{8}=\frac{3}{8}\left(8^{\frac{4}{3}}\right)=\frac{3}{8}\times 16=6\] Final answer: option (A) 6.
Exercise 10
If $\displaystyle f(x)=\int_{0}^{x} t \sin t d t$, then $\displaystyle f^{\prime}(x)$ is (A) $\displaystyle \cos x+x \sin x$ (B) $\displaystyle x \sin x$ (C) $\displaystyle x \cos x$ (D) $\displaystyle \sin x+x \cos x$
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NCERT’s answer
B
Apply the first fundamental theorem of calculus: if $\displaystyle f(x)=\displaystyle\int_{a}^{x}g(t)\,dt$ with $\displaystyle g$ continuous, then $\displaystyle f$ is differentiable and $\displaystyle f'(x)=g(x)$ — differentiating with respect to the upper limit simply replaces $\displaystyle t$ by $\displaystyle x$. Here $\displaystyle g(t)=t\sin t$, continuous for all $\displaystyle t$, so \[f'(x)=x\sin x\] The integral need not be evaluated at all; but as a check, integrating by parts gives $\displaystyle f(x)=\left[\sin t-t\cos t\right]_{0}^{x}=\sin x-x\cos x$, and then \[f'(x)=\cos x-\left(\cos x-x\sin x\right)=x\sin x\] the same result. Final answer: option (B) $\displaystyle x\sin x$.

EXERCISE 7.10 1–10 (part 22 of 27)

By using the properties of definite integrals, evaluate the integrals in Exercises $\displaystyle 1$ to 19.
Exercise 1
$\displaystyle \int_{0}^{\frac{\pi}{2}} \cos ^{2} x d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
Use the power-reduction identity $\displaystyle \cos^{2}x=\frac{1+\cos 2x}{2}$, since $\displaystyle \cos^{2}x$ has no elementary antiderivative in that form.\[\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{1}{2}\int_{0}^{\pi/2}\left(1+\cos 2x\right)dx=\frac{1}{2}\left[x+\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]At $\displaystyle x=\frac{\pi}{2}$ the bracket is $\displaystyle \frac{\pi}{2}+\frac{\sin\pi}{2}=\frac{\pi}{2}$; at $\displaystyle x=0$ it is $\displaystyle 0$.\[=\frac{1}{2}\left(\frac{\pi}{2}-0\right)=\frac{\pi}{4}.\]Hence $\displaystyle \displaystyle\int_{0}^{\pi/2}\cos^{2}x\,dx=\frac{\pi}{4}$.
Exercise 2
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}} d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
Apply the property $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$.Let \[I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx.\]Replacing $\displaystyle x$ by $\displaystyle \frac{\pi}{2}-x$ interchanges sine and cosine, because $\displaystyle \sin\left(\frac{\pi}{2}-x\right)=\cos x$ and $\displaystyle \cos\left(\frac{\pi}{2}-x\right)=\sin x$: \[I=\int_{0}^{\pi/2}\frac{\sqrt{\cos x}}{\sqrt{\cos x}+\sqrt{\sin x}}\,dx.\]The two forms have the same denominator, so adding them adds only the numerators: \[2I=\int_{0}^{\pi/2}\frac{\sqrt{\sin x}+\sqrt{\cos x}}{\sqrt{\sin x}+\sqrt{\cos x}}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence $\displaystyle I=\dfrac{\pi}{4}$.
Exercise 3
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin ^{\frac{3}{2}} x d x}{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
Same structure as the previous question: use $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$.Let \[I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx.\]Under $\displaystyle x\mapsto\frac{\pi}{2}-x$, $\displaystyle \sin x$ and $\displaystyle \cos x$ swap, so \[I=\int_{0}^{\pi/2}\frac{\cos^{3/2}x}{\cos^{3/2}x+\sin^{3/2}x}\,dx.\]The denominator is unchanged, so adding the two expressions gives \[2I=\int_{0}^{\pi/2}\frac{\sin^{3/2}x+\cos^{3/2}x}{\sin^{3/2}x+\cos^{3/2}x}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence $\displaystyle I=\dfrac{\pi}{4}$. (Notice the exponent never mattered — only that the two terms swap.)
Exercise 4
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{5} x d x}{\sin ^{5} x+\cos ^{5} x}$
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NCERT’s answer
$\displaystyle \frac{\pi}{4}$
Use $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$.Let \[I=\int_{0}^{\pi/2}\frac{\cos^{5}x}{\sin^{5}x+\cos^{5}x}\,dx.\]Replacing $\displaystyle x$ by $\displaystyle \frac{\pi}{2}-x$ swaps $\displaystyle \sin x$ and $\displaystyle \cos x$: \[I=\int_{0}^{\pi/2}\frac{\sin^{5}x}{\cos^{5}x+\sin^{5}x}\,dx.\]Adding the two (common denominator): \[2I=\int_{0}^{\pi/2}\frac{\cos^{5}x+\sin^{5}x}{\sin^{5}x+\cos^{5}x}\,dx=\int_{0}^{\pi/2}1\,dx=\frac{\pi}{2}.\]Hence $\displaystyle I=\dfrac{\pi}{4}$.
Exercise 5
$\displaystyle \int_{-5}^{5}|x+2| d x$
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NCERT’s answer
$\displaystyle 29$
The modulus changes its rule where $\displaystyle x+2=0$, i.e. at $\displaystyle x=-2$, and $\displaystyle -2$ lies inside $\displaystyle [-5,5]$. So split the integral there (additivity over adjacent intervals) — integrating $\displaystyle |x+2|$ in one piece is the standard mistake here.On $\displaystyle -5\le x\le-2$ we have $\displaystyle x+2\le 0$, so $\displaystyle |x+2|=-(x+2)$; on $\displaystyle -2\le x\le 5$ we have $\displaystyle x+2\ge 0$, so $\displaystyle |x+2|=x+2$.\[\int_{-5}^{5}|x+2|\,dx=-\int_{-5}^{-2}(x+2)\,dx+\int_{-2}^{5}(x+2)\,dx.\]First piece: \[-\left[\frac{x^{2}}{2}+2x\right]_{-5}^{-2}=-\left[(2-4)-\left(\frac{25}{2}-10\right)\right]=-\left[-2-\frac{5}{2}\right]=\frac{9}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}+2x\right]_{-2}^{5}=\left(\frac{25}{2}+10\right)-(2-4)=\frac{45}{2}+2=\frac{49}{2}.\]\[\int_{-5}^{5}|x+2|\,dx=\frac{9}{2}+\frac{49}{2}=29.\]
Exercise 6
$\displaystyle \int_{2}^{8}|x-5| d x$
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NCERT’s answer
$\displaystyle 9$
$\displaystyle |x-5|$ changes rule at $\displaystyle x=5$, which lies inside $\displaystyle [2,8]$, so split the integral at $\displaystyle 5$.On $\displaystyle 2\le x\le 5$, $\displaystyle x-5\le 0$ so $\displaystyle |x-5|=5-x$; on $\displaystyle 5\le x\le 8$, $\displaystyle |x-5|=x-5$.\[\int_{2}^{8}|x-5|\,dx=\int_{2}^{5}(5-x)\,dx+\int_{5}^{8}(x-5)\,dx.\]First piece: \[\left[5x-\frac{x^{2}}{2}\right]_{2}^{5}=\left(25-\frac{25}{2}\right)-(10-2)=\frac{25}{2}-8=\frac{9}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}-5x\right]_{5}^{8}=(32-40)-\left(\frac{25}{2}-25\right)=-8+\frac{25}{2}=\frac{9}{2}.\]\[\int_{2}^{8}|x-5|\,dx=\frac{9}{2}+\frac{9}{2}=9.\]
Exercise 7
$\displaystyle \int_{0}^{1} x(1-x)^{n} d x$
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NCERT’s answer
$\displaystyle \frac{1}{(n+1)(n+2)}$
Expanding $\displaystyle (1-x)^{n}$ by the binomial theorem works but is clumsy. Use instead $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=1$, which moves the power onto $\displaystyle x$.\[I=\int_{0}^{1}x(1-x)^{n}dx=\int_{0}^{1}(1-x)\bigl(1-(1-x)\bigr)^{n}dx=\int_{0}^{1}(1-x)\,x^{n}\,dx.\]Now it is a two-term power integral: \[I=\int_{0}^{1}\left(x^{n}-x^{n+1}\right)dx=\left[\frac{x^{n+1}}{n+1}-\frac{x^{n+2}}{n+2}\right]_{0}^{1}=\frac{1}{n+1}-\frac{1}{n+2}.\]Combining over a common denominator, \[I=\frac{(n+2)-(n+1)}{(n+1)(n+2)}=\frac{1}{(n+1)(n+2)}.\]Hence $\displaystyle \displaystyle\int_{0}^{1}x(1-x)^{n}dx=\frac{1}{(n+1)(n+2)}$ (for $\displaystyle n$ a non-negative integer, and indeed for any $\displaystyle n>-1$).
Exercise 8
$\displaystyle \int_{0}^{\frac{\pi}{4}} \log (1+\tan x) d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{8} \log 2$
Let $\displaystyle I=\displaystyle\int_{0}^{\pi/4}\log(1+\tan x)\,dx$ and apply $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{4}$.The key step is the tangent-difference formula: \[\tan\left(\frac{\pi}{4}-x\right)=\frac{\tan\dfrac{\pi}{4}-\tan x}{1+\tan\dfrac{\pi}{4}\tan x}=\frac{1-\tan x}{1+\tan x}.\]Therefore \[1+\tan\left(\frac{\pi}{4}-x\right)=1+\frac{1-\tan x}{1+\tan x}=\frac{(1+\tan x)+(1-\tan x)}{1+\tan x}=\frac{2}{1+\tan x}.\]So \[I=\int_{0}^{\pi/4}\log\left(\frac{2}{1+\tan x}\right)dx=\int_{0}^{\pi/4}\log 2\,dx-\int_{0}^{\pi/4}\log(1+\tan x)\,dx=\frac{\pi}{4}\log 2-I.\]Hence $\displaystyle 2I=\dfrac{\pi}{4}\log 2$, i.e. \[I=\frac{\pi}{8}\log 2.\]
Exercise 9
$\displaystyle \int_{0}^{2} x \sqrt{2-x} d x$
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NCERT’s answer
$\displaystyle \frac{16 \sqrt{2}}{15}$
Substitution $\displaystyle t=2-x$ works, and so does the property $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=2$ — take the second, which moves the awkward surd onto $\displaystyle x$ alone.\[I=\int_{0}^{2}x\sqrt{2-x}\,dx=\int_{0}^{2}(2-x)\sqrt{x}\,dx=\int_{0}^{2}\left(2x^{1/2}-x^{3/2}\right)dx.\]\[I=\left[2\cdot\frac{x^{3/2}}{3/2}-\frac{x^{5/2}}{5/2}\right]_{0}^{2}=\left[\frac{4}{3}x^{3/2}-\frac{2}{5}x^{5/2}\right]_{0}^{2}.\]With $\displaystyle 2^{3/2}=2\sqrt{2}$ and $\displaystyle 2^{5/2}=4\sqrt{2}$, \[I=\frac{4}{3}\left(2\sqrt{2}\right)-\frac{2}{5}\left(4\sqrt{2}\right)=\frac{8\sqrt{2}}{3}-\frac{8\sqrt{2}}{5}=8\sqrt{2}\left(\frac{5-3}{15}\right)=\frac{16\sqrt{2}}{15}.\]Hence $\displaystyle \displaystyle\int_{0}^{2}x\sqrt{2-x}\,dx=\frac{16\sqrt{2}}{15}$.
Exercise 10
$\displaystyle \int_{0}^{\frac{\pi}{2}}(2 \log \sin x-\log \sin 2 x) d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{2} \log \frac{1}{2}$
Simplify the integrand first, using $\displaystyle \sin 2x=2\sin x\cos x$ and $\displaystyle \log(abc)=\log a+\log b+\log c$: \[\log\sin 2x=\log 2+\log\sin x+\log\cos x.\]So \[2\log\sin x-\log\sin 2x=2\log\sin x-\log 2-\log\sin x-\log\cos x=\log\sin x-\log\cos x-\log 2.\]Hence \[I=\int_{0}^{\pi/2}\log\sin x\,dx-\int_{0}^{\pi/2}\log\cos x\,dx-\int_{0}^{\pi/2}\log 2\,dx.\]By $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$, the substitution $\displaystyle x\mapsto\frac{\pi}{2}-x$ turns $\displaystyle \log\sin x$ into $\displaystyle \log\cos x$, so \[\int_{0}^{\pi/2}\log\sin x\,dx=\int_{0}^{\pi/2}\log\cos x\,dx,\] and the first two integrals cancel exactly. (Each is a convergent improper integral, so the cancellation is legitimate.)What remains is \[I=-\log 2\int_{0}^{\pi/2}dx=-\frac{\pi}{2}\log 2.\]

EXERCISE 7.10 11–21 (part 23 of 27)

By using the properties of definite integrals, evaluate the integrals in Exercises $\displaystyle 1$ to 19.
Exercise 11
$\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{2} x d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{2}$
The limits are symmetric about $\displaystyle 0$, so test the parity of the integrand. Since $\displaystyle \sin(-x)=-\sin x$, \[\sin^{2}(-x)=(-\sin x)^{2}=\sin^{2}x,\] so $\displaystyle \sin^{2}x$ is even, and $\displaystyle \displaystyle\int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx$.\[I=2\int_{0}^{\pi/2}\sin^{2}x\,dx=2\cdot\frac{1}{2}\int_{0}^{\pi/2}(1-\cos 2x)\,dx=\left[x-\frac{\sin 2x}{2}\right]_{0}^{\pi/2}.\]\[I=\left(\frac{\pi}{2}-\frac{\sin\pi}{2}\right)-(0-0)=\frac{\pi}{2}.\]Hence $\displaystyle \displaystyle\int_{-\pi/2}^{\pi/2}\sin^{2}x\,dx=\frac{\pi}{2}$.
Exercise 12
$\displaystyle \int_{0}^{\pi} \frac{x d x}{1+\sin x}$
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NCERT’s answer
$\displaystyle \pi$
Let $\displaystyle I=\displaystyle\int_{0}^{\pi}\frac{x}{1+\sin x}\,dx$ and use $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\pi$. Since $\displaystyle \sin(\pi-x)=\sin x$, only the numerator changes: \[I=\int_{0}^{\pi}\frac{\pi-x}{1+\sin x}\,dx.\]Adding the two expressions kills the $\displaystyle x$: \[2I=\int_{0}^{\pi}\frac{x+(\pi-x)}{1+\sin x}\,dx=\pi\int_{0}^{\pi}\frac{dx}{1+\sin x}.\]Rationalise the denominator by multiplying above and below by $\displaystyle 1-\sin x$: \[\frac{1}{1+\sin x}=\frac{1-\sin x}{1-\sin^{2}x}=\frac{1-\sin x}{\cos^{2}x}=\sec^{2}x-\sec x\tan x.\]An antiderivative is $\displaystyle \tan x-\sec x$. This is the step to be careful with: $\displaystyle \tan x$ and $\displaystyle \sec x$ each blow up at $\displaystyle x=\frac{\pi}{2}$, which is inside $\displaystyle [0,\pi]$, so rewrite the antiderivative in a form that is continuous there: \[\tan x-\sec x=\frac{\sin x-1}{\cos x}=\frac{(\sin x-1)(\sin x+1)}{\cos x(1+\sin x)}=\frac{-\cos^{2}x}{\cos x(1+\sin x)}=\frac{-\cos x}{1+\sin x}.\]This last expression is continuous on all of $\displaystyle [0,\pi]$, so it may be used at the limits: \[\int_{0}^{\pi}\frac{dx}{1+\sin x}=\left[\frac{-\cos x}{1+\sin x}\right]_{0}^{\pi}=\frac{-\cos\pi}{1+0}-\frac{-\cos 0}{1+0}=1+1=2.\]Hence $\displaystyle 2I=\pi\cdot 2$, so \[I=\pi.\]
Exercise 13
$\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \sin ^{7} x d x$
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NCERT’s answer
$\displaystyle 0$
The limits are symmetric about $\displaystyle 0$, so check parity. Since $\displaystyle \sin(-x)=-\sin x$, \[\sin^{7}(-x)=(-\sin x)^{7}=-\sin^{7}x,\] an odd power of an odd function, so $\displaystyle f(x)=\sin^{7}x$ is odd.By the property $\displaystyle \displaystyle\int_{-a}^{a}f(x)\,dx=0$ when $\displaystyle f$ is odd, \[\int_{-\pi/2}^{\pi/2}\sin^{7}x\,dx=0.\]Hence the value is $\displaystyle 0$.
Exercise 14
$\displaystyle \int_{0}^{2 \pi} \cos ^{5} x d x$
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NCERT’s answer
$\displaystyle 0$
Use $\displaystyle \displaystyle\int_{0}^{2a}f(x)\,dx=\int_{0}^{a}f(x)\,dx+\int_{0}^{a}f(2a-x)\,dx$ with $\displaystyle a=\pi$.Since $\displaystyle \cos(2\pi-x)=\cos x$, we get $\displaystyle f(2\pi-x)=\cos^{5}x=f(x)$, so \[\int_{0}^{2\pi}\cos^{5}x\,dx=2\int_{0}^{\pi}\cos^{5}x\,dx.\]Now apply the same idea on $\displaystyle [0,\pi]$ with $\displaystyle a=\frac{\pi}{2}$. Here $\displaystyle \cos(\pi-x)=-\cos x$, so $\displaystyle \cos^{5}(\pi-x)=-\cos^{5}x$: \[\int_{0}^{\pi}\cos^{5}x\,dx=\int_{0}^{\pi/2}\left[\cos^{5}x+\cos^{5}(\pi-x)\right]dx=\int_{0}^{\pi/2}\left[\cos^{5}x-\cos^{5}x\right]dx=0.\]Hence \[\int_{0}^{2\pi}\cos^{5}x\,dx=2\cdot 0=0.\]
Exercise 15
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\sin x-\cos x}{1+\sin x \cos x} d x$
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NCERT’s answer
$\displaystyle 0$
Let \[I=\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx\] and apply $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$.Under $\displaystyle x\mapsto\frac{\pi}{2}-x$, $\displaystyle \sin x$ and $\displaystyle \cos x$ swap. The denominator $\displaystyle 1+\sin x\cos x$ is symmetric in them, so it is unchanged; the numerator reverses sign: \[I=\int_{0}^{\pi/2}\frac{\cos x-\sin x}{1+\sin x\cos x}\,dx=-\int_{0}^{\pi/2}\frac{\sin x-\cos x}{1+\sin x\cos x}\,dx=-I.\]Hence $\displaystyle 2I=0$, so \[I=0.\]
Exercise 16
$\displaystyle \int_{0}^{\pi} \log (1+\cos x) d x$
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NCERT’s answer
$\displaystyle -\pi \log 2$
Let $\displaystyle I=\displaystyle\int_{0}^{\pi}\log(1+\cos x)\,dx$ and apply $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\pi$. Since $\displaystyle \cos(\pi-x)=-\cos x$, \[I=\int_{0}^{\pi}\log(1-\cos x)\,dx.\]Add the two forms and use $\displaystyle \log A+\log B=\log AB$: \[2I=\int_{0}^{\pi}\log\left[(1+\cos x)(1-\cos x)\right]dx=\int_{0}^{\pi}\log\left(\sin^{2}x\right)dx=2\int_{0}^{\pi}\log\sin x\,dx.\]So everything reduces to the standard value of $\displaystyle \displaystyle\int_{0}^{\pi/2}\log\sin x\,dx$. Call it $\displaystyle A$. By the same reflection property, $\displaystyle A=\displaystyle\int_{0}^{\pi/2}\log\cos x\,dx$, so \[2A=\int_{0}^{\pi/2}\log(\sin x\cos x)\,dx=\int_{0}^{\pi/2}\log\left(\frac{\sin 2x}{2}\right)dx=\int_{0}^{\pi/2}\log\sin 2x\,dx-\frac{\pi}{2}\log 2.\]In the remaining integral put $\displaystyle t=2x$, $\displaystyle dt=2\,dx$; the limits become $\displaystyle 0$ to $\displaystyle \pi$: \[\int_{0}^{\pi/2}\log\sin 2x\,dx=\frac{1}{2}\int_{0}^{\pi}\log\sin t\,dt=\frac{1}{2}\left(2A\right)=A,\] where $\displaystyle \displaystyle\int_{0}^{\pi}\log\sin t\,dt=2A$ because $\displaystyle \sin(\pi-t)=\sin t$ makes the graph symmetric about $\displaystyle t=\frac{\pi}{2}$.Therefore $\displaystyle 2A=A-\frac{\pi}{2}\log 2$, giving $\displaystyle A=-\frac{\pi}{2}\log 2$ and $\displaystyle \displaystyle\int_{0}^{\pi}\log\sin x\,dx=2A=-\pi\log 2$.Substituting back, \[2I=2(-\pi\log 2)\quad\Longrightarrow\quad I=-\pi\log 2.\]Hence $\displaystyle \displaystyle\int_{0}^{\pi}\log(1+\cos x)\,dx=-\pi\log 2$.
Exercise 17
$\displaystyle \int_{0}^{a} \frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}} d x$
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NCERT’s answer
$\displaystyle \frac{a}{2}$
Apply $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$. Let \[I=\int_{0}^{a}\frac{\sqrt{x}}{\sqrt{x}+\sqrt{a-x}}\,dx.\]Replacing $\displaystyle x$ by $\displaystyle a-x$ interchanges $\displaystyle \sqrt{x}$ and $\displaystyle \sqrt{a-x}$: \[I=\int_{0}^{a}\frac{\sqrt{a-x}}{\sqrt{a-x}+\sqrt{x}}\,dx.\]The denominators agree, so adding the two expressions gives \[2I=\int_{0}^{a}\frac{\sqrt{x}+\sqrt{a-x}}{\sqrt{x}+\sqrt{a-x}}\,dx=\int_{0}^{a}1\,dx=a.\]Hence $\displaystyle I=\dfrac{a}{2}$ (for $\displaystyle a>0$).
Exercise 18
$\displaystyle \int_{0}^{4}|x-1| d x$
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NCERT’s answer
$\displaystyle 5$
$\displaystyle |x-1|$ changes rule at $\displaystyle x=1$, which lies inside $\displaystyle [0,4]$, so split the integral there.On $\displaystyle 0\le x\le 1$, $\displaystyle x-1\le 0$ so $\displaystyle |x-1|=1-x$; on $\displaystyle 1\le x\le 4$, $\displaystyle |x-1|=x-1$.\[\int_{0}^{4}|x-1|\,dx=\int_{0}^{1}(1-x)\,dx+\int_{1}^{4}(x-1)\,dx.\]First piece: \[\left[x-\frac{x^{2}}{2}\right]_{0}^{1}=\left(1-\frac{1}{2}\right)-0=\frac{1}{2}.\]Second piece: \[\left[\frac{x^{2}}{2}-x\right]_{1}^{4}=(8-4)-\left(\frac{1}{2}-1\right)=4+\frac{1}{2}=\frac{9}{2}.\]\[\int_{0}^{4}|x-1|\,dx=\frac{1}{2}+\frac{9}{2}=5.\]
Exercise 19
Show that $\displaystyle \int_{0}^{a} f(x) g(x) d x=2 \int_{0}^{a} f(x) d x$, if $\displaystyle f$ and $\displaystyle g$ are defined as $\displaystyle f(x)=f(a-x)$ and $\displaystyle g(x)+g(a-x)=4$
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Write $\displaystyle I=\displaystyle\int_{0}^{a}f(x)g(x)\,dx$ and apply the property $\displaystyle \displaystyle\int_{0}^{a}h(x)\,dx=\int_{0}^{a}h(a-x)\,dx$ to the whole integrand $\displaystyle h=fg$: \[I=\int_{0}^{a}f(a-x)\,g(a-x)\,dx.\]Now use the two given conditions. First $\displaystyle f(a-x)=f(x)$. Second, $\displaystyle g(x)+g(a-x)=4$ gives $\displaystyle g(a-x)=4-g(x)$. Substituting both: \[I=\int_{0}^{a}f(x)\left[4-g(x)\right]dx=4\int_{0}^{a}f(x)\,dx-\int_{0}^{a}f(x)g(x)\,dx=4\int_{0}^{a}f(x)\,dx-I.\]The integral $\displaystyle I$ has reappeared, so bring it across: \[2I=4\int_{0}^{a}f(x)\,dx\quad\Longrightarrow\quad I=2\int_{0}^{a}f(x)\,dx.\]That is, $\displaystyle \displaystyle\int_{0}^{a}f(x)g(x)\,dx=2\int_{0}^{a}f(x)\,dx$, as required.
Choose the correct answer in Exercises $\displaystyle 20$ and 21.
Exercise 20
The value of $\displaystyle \int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}\left(x^{3}+x \cos x+\tan ^{5} x+1\right) d x$ is (A) $\displaystyle 0$ (B) $\displaystyle 2$ (C) $\displaystyle \pi$ (D) $\displaystyle 1$
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NCERT’s answer
C
The limits $\displaystyle -\frac{\pi}{2}$ and $\displaystyle \frac{\pi}{2}$ are symmetric about $\displaystyle 0$, so split the integrand by parity and use \[\int_{-a}^{a}f(x)\,dx=0\ (f\ \text{odd}),\qquad \int_{-a}^{a}f(x)\,dx=2\int_{0}^{a}f(x)\,dx\ (f\ \text{even}).\]Check each term:
$\displaystyle (-x)^{3}=-x^{3}$: $\displaystyle x^{3}$ is odd.

$\displaystyle (-x)\cos(-x)=-x\cos x$ (cosine is even): $\displaystyle x\cos x$ is odd.

$\displaystyle \tan^{5}(-x)=(-\tan x)^{5}=-\tan^{5}x$: $\displaystyle \tan^{5}x$ is odd, and its contributions on the two halves cancel term-for-term.

$\displaystyle 1$ is even.
All three odd terms integrate to $\displaystyle 0$ over the symmetric interval, leaving only the constant: \[\int_{-\pi/2}^{\pi/2}1\,dx=\frac{\pi}{2}-\left(-\frac{\pi}{2}\right)=\pi.\]Hence the value is $\displaystyle \pi$ — option (C).
Exercise 21
The value of $\displaystyle \int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ is (A) $\displaystyle 2$ (B) $\displaystyle \frac{3}{4}$ (C) $\displaystyle 0$ (D) -$\displaystyle 2$
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NCERT’s answer
C
Let \[I=\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx\] and apply $\displaystyle \displaystyle\int_{0}^{a}f(x)\,dx=\int_{0}^{a}f(a-x)\,dx$ with $\displaystyle a=\frac{\pi}{2}$, which swaps $\displaystyle \sin x$ and $\displaystyle \cos x$: \[I=\int_{0}^{\pi/2}\log\left(\frac{4+3\cos x}{4+3\sin x}\right)dx.\]The new integrand is the logarithm of the reciprocal of the old one, and $\displaystyle \log\frac{1}{t}=-\log t$, so \[I=-\int_{0}^{\pi/2}\log\left(\frac{4+3\sin x}{4+3\cos x}\right)dx=-I.\]Hence $\displaystyle 2I=0$, so $\displaystyle I=0$ — option (C).

Miscellaneous Exercise 1–10 (part 24 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 1
$\displaystyle \frac{1}{x-x^{3}}$
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NCERT’s answer
$\displaystyle \frac{1}{2} \log \left|\frac{x^{2}}{1-x^{2}}\right|+\mathrm{C}$
Factorise the denominator and resolve into partial fractions: \[\frac{1}{x-x^{3}}=\frac{1}{x(1-x)(1+x)}=\frac{A}{x}+\frac{B}{1-x}+\frac{C}{1+x}. \] Clearing denominators, $\displaystyle 1=A(1-x)(1+x)+Bx(1+x)+Cx(1-x)$. Putting $\displaystyle x=0$ gives $\displaystyle A=1$; $\displaystyle x=1$ gives $\displaystyle 1=2B$, so $\displaystyle B=\frac12$; $\displaystyle x=-1$ gives $\displaystyle 1=-2C$, so $\displaystyle C=-\frac12$. Hence \[\int\frac{dx}{x-x^{3}}=\int\frac{dx}{x}+\frac12\int\frac{dx}{1-x}-\frac12\int\frac{dx}{1+x}. \] The middle integral carries a sign from $\displaystyle \frac{d}{dx}(1-x)=-1$, so $\displaystyle \int\frac{dx}{1-x}=-\log|1-x|$. Therefore \[\int\frac{dx}{x-x^{3}}=\log|x|-\frac12\log|1-x|-\frac12\log|1+x|+\mathrm{C}=\log|x|-\frac12\log\left|1-x^{2}\right|+\mathrm{C}. \]
Exercise 2
$\displaystyle \frac{1}{\sqrt{x+a}+\sqrt{x+b}}$
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NCERT’s answer
$\displaystyle \frac{2}{3(a-b)}\left[(x+a)^{\frac{3}{2}}-(x+b)^{\frac{3}{2}}\right]+\mathrm{C}$
Rationalise, because $\displaystyle \left(\sqrt{x+a}\right)^{2}-\left(\sqrt{x+b}\right)^{2}=a-b$ is a constant (this needs $\displaystyle a\neq b$): \[\frac{1}{\sqrt{x+a}+\sqrt{x+b}}=\frac{\sqrt{x+a}-\sqrt{x+b}}{(x+a)-(x+b)}=\frac{\sqrt{x+a}-\sqrt{x+b}}{a-b}. \] Now integrate term by term using $\displaystyle \int (x+k)^{1/2}dx=\frac23 (x+k)^{3/2}$: \[\int\frac{dx}{\sqrt{x+a}+\sqrt{x+b}}=\frac{1}{a-b}\left[\frac23 (x+a)^{3/2}-\frac23 (x+b)^{3/2}\right]+\mathrm{C} \] \[=\frac{2}{3(a-b)}\left[(x+a)^{3/2}-(x+b)^{3/2}\right]+\mathrm{C}. \]
Exercise 3
$\displaystyle \frac{1}{x \sqrt{a x-x^{2}}}$ [Hint:Put $\displaystyle x=\frac{a}{t}$ ]
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NCERT’s answer
$\displaystyle -\frac{2}{a} \sqrt{\frac{(a-x)}{x}}+\mathrm{C}$
Follow the hint and put $\displaystyle x=\dfrac{a}{t}$, so $\displaystyle dx=-\dfrac{a}{t^{2}}dt$. Then \[ax-x^{2}=\frac{a^{2}}{t}-\frac{a^{2}}{t^{2}}=\frac{a^{2}(t-1)}{t^{2}},\qquad \sqrt{ax-x^{2}}=\frac{a}{t}\sqrt{t-1}, \] where $\displaystyle t>0$ because the integrand is only defined for $\displaystyle 0<x<a$. Hence \[x\sqrt{ax-x^{2}}=\frac{a}{t}\cdot\frac{a}{t}\sqrt{t-1}=\frac{a^{2}}{t^{2}}\sqrt{t-1}, \] and the integral becomes \[\int\frac{dx}{x\sqrt{ax-x^{2}}}=\int\frac{-\dfrac{a}{t^{2}}\,dt}{\dfrac{a^{2}}{t^{2}}\sqrt{t-1}}=-\frac{1}{a}\int\frac{dt}{\sqrt{t-1}}=-\frac{2}{a}\sqrt{t-1}+\mathrm{C}. \] Back-substituting $\displaystyle t=\dfrac{a}{x}$, \[=-\frac{2}{a}\sqrt{\frac{a}{x}-1}+\mathrm{C}=-\frac{2}{a}\sqrt{\frac{a-x}{x}}+\mathrm{C}. \]
Exercise 4
$\displaystyle \frac{1}{x^{2}\left(x^{4}+1\right)^{\frac{3}{4}}}$
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NCERT’s answer
$\displaystyle -\left(1+\frac{1}{x^{4}}\right)^{\frac{1}{4}}+\mathrm{C}$
The step that makes this work is taking $\displaystyle x^{4}$ out of the bracket before substituting (take $\displaystyle x>0$): \[\left(x^{4}+1\right)^{3/4}=\left[x^{4}\left(1+x^{-4}\right)\right]^{3/4}=x^{3}\left(1+\frac{1}{x^{4}}\right)^{3/4}. \] So the integrand is \[\frac{1}{x^{2}\left(x^{4}+1\right)^{3/4}}=\frac{1}{x^{5}}\left(1+\frac{1}{x^{4}}\right)^{-3/4}. \] Put $\displaystyle t=1+\dfrac{1}{x^{4}}$, so $\displaystyle dt=-\dfrac{4}{x^{5}}dx$, i.e. $\displaystyle \dfrac{dx}{x^{5}}=-\dfrac{dt}{4}$: \[\int\frac{dx}{x^{2}\left(x^{4}+1\right)^{3/4}}=-\frac14\int t^{-3/4}dt=-\frac14\cdot 4\,t^{1/4}+\mathrm{C}=-t^{1/4}+\mathrm{C}. \] \[=-\left(1+\frac{1}{x^{4}}\right)^{1/4}+\mathrm{C}. \]
Exercise 5
$\displaystyle \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}$ [Hint: $\displaystyle \frac{1}{x^{\frac{1}{2}}+x^{\frac{1}{3}}}=\frac{1}{x^{\frac{1}{3}}\left(1+x^{\frac{1}{6}}\right)}$, put $\displaystyle x=t^{6}$ ]
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NCERT’s answer
$\displaystyle 2 \sqrt{x}-3 x^{\frac{1}{3}}+6 x^{\frac{1}{6}}-6 \log \left(1+x^{\frac{1}{6}}\right)+\mathrm{C}$
Both fractional powers are cleared at once by $\displaystyle x=t^{6}$, since $\displaystyle 6$ is the l.c.m. of $\displaystyle 2$ and $\displaystyle 3$. Then $\displaystyle dx=6t^{5}dt$, $\displaystyle x^{1/2}=t^{3}$, $\displaystyle x^{1/3}=t^{2}$: \[\int\frac{dx}{x^{1/2}+x^{1/3}}=\int\frac{6t^{5}}{t^{3}+t^{2}}dt=\int\frac{6t^{5}}{t^{2}(t+1)}dt=6\int\frac{t^{3}}{t+1}dt. \] This rational function is improper, so divide before integrating: $\displaystyle \dfrac{t^{3}}{t+1}=t^{2}-t+1-\dfrac{1}{t+1}$. Hence \[6\int\left(t^{2}-t+1-\frac{1}{t+1}\right)dt=2t^{3}-3t^{2}+6t-6\log|t+1|+\mathrm{C}. \] With $\displaystyle t=x^{1/6}$ (so $\displaystyle t>0$), \[=2\sqrt{x}-3x^{1/3}+6x^{1/6}-6\log\left(1+x^{1/6}\right)+\mathrm{C}. \]
Exercise 6
$\displaystyle \frac{5 x}{(x+1)\left(x^{2}+9\right)}$
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NCERT’s answer
$\displaystyle -\frac{1}{2} \log |x+1|+\frac{1}{4} \log \left(x^{2}+9\right)+\frac{3}{2} \tan ^{-1} \frac{x}{3}+\mathrm{C}$
The denominator has a linear factor and an irreducible quadratic, so the partial-fraction form must be \[\frac{5x}{(x+1)\left(x^{2}+9\right)}=\frac{A}{x+1}+\frac{Bx+C}{x^{2}+9}, \] giving $\displaystyle 5x=A\left(x^{2}+9\right)+(Bx+C)(x+1)$. Putting $\displaystyle x=-1$: $\displaystyle -5=10A$, so $\displaystyle A=-\frac12$. Comparing coefficients of $\displaystyle x^{2}$: $\displaystyle 0=A+B$, so $\displaystyle B=\frac12$. Comparing constants: $\displaystyle 0=9A+C$, so $\displaystyle C=\frac92$. \[\int\frac{5x\,dx}{(x+1)\left(x^{2}+9\right)}=-\frac12\int\frac{dx}{x+1}+\frac12\int\frac{x\,dx}{x^{2}+9}+\frac92\int\frac{dx}{x^{2}+9}. \] Using $\displaystyle \int\frac{x\,dx}{x^{2}+9}=\frac12\log\left(x^{2}+9\right)$ and $\displaystyle \int\frac{dx}{x^{2}+9}=\frac13\tan^{-1}\frac{x}{3}$, \[=-\frac12\log|x+1|+\frac14\log\left(x^{2}+9\right)+\frac32\tan^{-1}\frac{x}{3}+\mathrm{C}. \]
Exercise 7
$\displaystyle \frac{\sin x}{\sin (x-a)}$
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NCERT’s answer
$\displaystyle \sin a \log |\sin (x-a)|+x \cos a+\mathrm{C}$
Numerator and denominator have different angles, so rewrite the numerator's angle in terms of the denominator's, $\displaystyle x=(x-a)+a$, and expand by the compound-angle formula: \[\sin x=\sin\big((x-a)+a\big)=\sin (x-a)\cos a+\cos (x-a)\sin a. \] Dividing by $\displaystyle \sin (x-a)$, \[\frac{\sin x}{\sin (x-a)}=\cos a+\sin a\,\cot (x-a). \] Here $\displaystyle a$ is a constant, so \[\int\frac{\sin x}{\sin (x-a)}dx=\cos a\int dx+\sin a\int\cot (x-a)\,dx=x\cos a+\sin a\,\log|\sin (x-a)|+\mathrm{C}. \]
Exercise 8
$\displaystyle \frac{e^{5 \log x}-e^{4 \log x}}{e^{3 \log x}-e^{2 \log x}}$
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NCERT’s answer
$\displaystyle \frac{x^{3}}{3}+C$
Simplify before integrating, using $\displaystyle e^{n\log x}=e^{\log x^{n}}=x^{n}$, valid for $\displaystyle x>0$: \[\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}=\frac{x^{5}-x^{4}}{x^{3}-x^{2}}=\frac{x^{4}(x-1)}{x^{2}(x-1)}=x^{2},\qquad x>0,\;x\neq 1. \] Hence \[\int\frac{e^{5\log x}-e^{4\log x}}{e^{3\log x}-e^{2\log x}}\,dx=\int x^{2}dx=\frac{x^{3}}{3}+\mathrm{C}. \]
Exercise 9
$\displaystyle \frac{\cos x}{\sqrt{4-\sin ^{2} x}}$
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NCERT’s answer
$\displaystyle \sin ^{-1}\left(\frac{\sin x}{2}\right)+\mathrm{C}$
The numerator is exactly the derivative of $\displaystyle \sin x$, so substitute $\displaystyle t=\sin x$, $\displaystyle dt=\cos x\,dx$: \[\int\frac{\cos x}{\sqrt{4-\sin^{2}x}}dx=\int\frac{dt}{\sqrt{2^{2}-t^{2}}}. \] By the standard form $\displaystyle \int\frac{dt}{\sqrt{a^{2}-t^{2}}}=\sin^{-1}\frac{t}{a}+\mathrm{C}$, \[=\sin^{-1}\frac{t}{2}+\mathrm{C}=\sin^{-1}\left(\frac{\sin x}{2}\right)+\mathrm{C}. \]
Exercise 10
$\displaystyle \frac{\sin ^{8}-\cos ^{8} x}{1-2 \sin ^{2} x \cos ^{2} x}$
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NCERT’s answer
$\displaystyle -\frac{1}{2} \sin 2 x+C$
Factorise the numerator as a difference of squares twice, using $\displaystyle \sin^{2}x+\cos^{2}x=1$: \[\sin^{8}x-\cos^{8}x=\left(\sin^{4}x-\cos^{4}x\right)\left(\sin^{4}x+\cos^{4}x\right)=\left(\sin^{2}x-\cos^{2}x\right)\left(\sin^{4}x+\cos^{4}x\right). \] The denominator is precisely that second factor: \[1-2\sin^{2}x\cos^{2}x=\left(\sin^{2}x+\cos^{2}x\right)^{2}-2\sin^{2}x\cos^{2}x=\sin^{4}x+\cos^{4}x. \] So the integrand reduces to $\displaystyle \sin^{2}x-\cos^{2}x=-\cos 2x$, and \[\int\frac{\sin^{8}x-\cos^{8}x}{1-2\sin^{2}x\cos^{2}x}dx=-\int\cos 2x\,dx=-\frac{\sin 2x}{2}+\mathrm{C}. \]

Miscellaneous Exercise 11–20 (part 25 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 11
$\displaystyle \frac{1}{\cos (x+a) \cos (x+b)}$
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NCERT’s answer
$\displaystyle \frac{1}{\sin (a-b)} \log \left|\frac{\cos (x+b)}{\cos (x+a)}\right|+\mathrm{C}$
Multiply and divide by the constant $\displaystyle \sin (a-b)$, written as the sine of the difference of the two angles appearing in the denominator: \[\sin (a-b)=\sin\big((x+a)-(x+b)\big)=\sin (x+a)\cos (x+b)-\cos (x+a)\sin (x+b). \] Dividing this identity by $\displaystyle \cos (x+a)\cos (x+b)$, \[\frac{1}{\cos (x+a)\cos (x+b)}=\frac{1}{\sin (a-b)}\big[\tan (x+a)-\tan (x+b)\big],\qquad a\neq b. \] Since $\displaystyle \int\tan u\,du=-\log|\cos u|$, \[\int\frac{dx}{\cos (x+a)\cos (x+b)}=\frac{1}{\sin (a-b)}\Big[-\log|\cos (x+a)|+\log|\cos (x+b)|\Big]+\mathrm{C} \] \[=\frac{1}{\sin (a-b)}\log\left|\frac{\cos (x+b)}{\cos (x+a)}\right|+\mathrm{C}. \]
Exercise 12
$\displaystyle \frac{x^{3}}{\sqrt{1-x^{8}}}$
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NCERT’s answer
$\displaystyle \frac{1}{4} \sin ^{-1}\left(x^{4}\right)+C$
Note that $\displaystyle x^{8}=\left(x^{4}\right)^{2}$ and $\displaystyle \frac{d}{dx}x^{4}=4x^{3}$, which is the numerator up to a constant. Put $\displaystyle t=x^{4}$, so $\displaystyle dt=4x^{3}dx$: \[\int\frac{x^{3}}{\sqrt{1-x^{8}}}dx=\frac14\int\frac{dt}{\sqrt{1-t^{2}}}=\frac14\sin^{-1}t+\mathrm{C}=\frac14\sin^{-1}\left(x^{4}\right)+\mathrm{C}. \]
Exercise 13
$\displaystyle \frac{e^{x}}{\left(1+e^{x}\right)\left(2+e^{x}\right)}$
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NCERT’s answer
$\displaystyle \log \left(\frac{1+e^{x}}{2+e^{x}}\right)+\mathrm{C}$
Put $\displaystyle t=e^{x}$, so $\displaystyle dt=e^{x}dx$ and the whole numerator is absorbed: \[\int\frac{e^{x}\,dx}{\left(1+e^{x}\right)\left(2+e^{x}\right)}=\int\frac{dt}{(1+t)(2+t)}. \] By partial fractions, $\displaystyle \dfrac{1}{(1+t)(2+t)}=\dfrac{1}{1+t}-\dfrac{1}{2+t}$. Hence \[=\log|1+t|-\log|2+t|+\mathrm{C}=\log\left|\frac{1+t}{2+t}\right|+\mathrm{C}=\log\left(\frac{1+e^{x}}{2+e^{x}}\right)+\mathrm{C}, \] the modulus signs being unnecessary because $\displaystyle 1+e^{x}$ and $\displaystyle 2+e^{x}$ are always positive.
Exercise 14
$\displaystyle \frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}$
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NCERT’s answer
$\displaystyle \frac{1}{3} \tan ^{-1} x-\frac{1}{6} \tan ^{-1} \frac{x}{2}+C$
Both factors are quadratic in $\displaystyle x^{2}$, so split them by writing $\displaystyle 1$ as their difference divided by $\displaystyle 3$: \[\frac{1}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac13\cdot\frac{\left(x^{2}+4\right)-\left(x^{2}+1\right)}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac13\left[\frac{1}{x^{2}+1}-\frac{1}{x^{2}+4}\right]. \] Using $\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac1a\tan^{-1}\frac{x}{a}$, \[\int\frac{dx}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac13\left[\tan^{-1}x-\frac12\tan^{-1}\frac{x}{2}\right]+\mathrm{C}=\frac13\tan^{-1}x-\frac16\tan^{-1}\frac{x}{2}+\mathrm{C}. \]
Exercise 15
$\displaystyle \cos ^{3} x e^{\log \sin x}$
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NCERT’s answer
$\displaystyle -\frac{1}{4} \cos ^{4} x+\mathrm{C}$
First simplify $\displaystyle e^{\log\sin x}=\sin x$ (valid where $\displaystyle \sin x>0$), so the integrand is $\displaystyle \cos^{3}x\sin x$. Put $\displaystyle t=\cos x$, so $\displaystyle dt=-\sin x\,dx$: \[\int\cos^{3}x\,\sin x\,dx=-\int t^{3}dt=-\frac{t^{4}}{4}+\mathrm{C}=-\frac{\cos^{4}x}{4}+\mathrm{C}. \]
Exercise 16
$\displaystyle e^{3 \log x}\left(x^{4}+1\right)^{-1}$
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NCERT’s answer
$\displaystyle \frac{1}{4} \log \left(x^{4}+1\right)+\mathrm{C}$
Since $\displaystyle e^{3\log x}=x^{3}$ for $\displaystyle x>0$, the integrand is $\displaystyle \dfrac{x^{3}}{x^{4}+1}$. Put $\displaystyle t=x^{4}+1$, so $\displaystyle dt=4x^{3}dx$: \[\int\frac{x^{3}}{x^{4}+1}dx=\frac14\int\frac{dt}{t}=\frac14\log|t|+\mathrm{C}=\frac14\log\left(x^{4}+1\right)+\mathrm{C}, \] no modulus being needed since $\displaystyle x^{4}+1>0$.
Exercise 17
$\displaystyle f^{\prime}(a x+b)[f(a x+b)]^{n}$
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NCERT’s answer
$\displaystyle \frac{[f(a x+b)]^{n+1}}{a(n+1)}+\mathrm{C}$
The derivative of the bracket is present, but only up to the factor $\displaystyle a$: by the chain rule $\displaystyle \dfrac{d}{dx}f(ax+b)=a\,f'(ax+b)$. Put $\displaystyle t=f(ax+b)$, so $\displaystyle dt=a\,f'(ax+b)\,dx$, i.e. $\displaystyle f'(ax+b)\,dx=\dfrac{dt}{a}$: \[\int f'(ax+b)\big[f(ax+b)\big]^{n}dx=\frac1a\int t^{n}dt=\frac{t^{n+1}}{a(n+1)}+\mathrm{C}. \] \[=\frac{\big[f(ax+b)\big]^{n+1}}{a(n+1)}+\mathrm{C},\qquad a\neq 0,\; n\neq -1. \] (For the excluded case $\displaystyle n=-1$ the answer is $\displaystyle \frac1a\log\big|f(ax+b)\big|+\mathrm{C}$.)
Exercise 18
$\displaystyle \frac{1}{\sqrt{\sin ^{3} x \sin (x+\alpha)}}$
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NCERT’s answer
$\displaystyle \frac{-2}{\sin \alpha} \sqrt{\frac{\sin (x+\alpha)}{\sin x}}+\mathrm{C}$
Expand $\displaystyle \sin (x+\alpha)=\sin x\cos\alpha+\cos x\sin\alpha$ and deliberately take $\displaystyle \sin x$ out of the root: \[\sin^{3}x\,\sin (x+\alpha)=\sin^{4}x\left[\cos\alpha+\cot x\,\sin\alpha\right],\qquad\text{so}\qquad \sqrt{\sin^{3}x\,\sin (x+\alpha)}=\sin^{2}x\sqrt{\cos\alpha+\sin\alpha\cot x}. \] The integrand therefore becomes $\displaystyle \dfrac{\mathrm{cosec}^{2}x}{\sqrt{\cos\alpha+\sin\alpha\cot x}}$. Put $\displaystyle t=\cos\alpha+\sin\alpha\cot x$, so $\displaystyle dt=-\sin\alpha\,\mathrm{cosec}^{2}x\,dx$: \[\int\frac{dx}{\sqrt{\sin^{3}x\,\sin (x+\alpha)}}=-\frac{1}{\sin\alpha}\int\frac{dt}{\sqrt{t}}=-\frac{2}{\sin\alpha}\sqrt{t}+\mathrm{C}. \] Since $\displaystyle \cos\alpha+\sin\alpha\cot x=\dfrac{\sin x\cos\alpha+\cos x\sin\alpha}{\sin x}=\dfrac{\sin (x+\alpha)}{\sin x}$, \[=-\frac{2}{\sin\alpha}\sqrt{\frac{\sin (x+\alpha)}{\sin x}}+\mathrm{C},\qquad \sin\alpha\neq 0. \]
Exercise 19
$\displaystyle \sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}$
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NCERT’s answer
$\displaystyle -2 \sqrt{1-x}+\cos ^{-1} \sqrt{x}+\sqrt{x-x^{2}}+\mathrm{C}$
Rationalise inside the root by multiplying numerator and denominator by $\displaystyle 1-\sqrt{x}$: \[\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}=\sqrt{\frac{\left(1-\sqrt{x}\right)^{2}}{1-x}}=\frac{1-\sqrt{x}}{\sqrt{1-x}},\qquad 0\le x<1, \] where $\displaystyle 1-\sqrt{x}\ge 0$ on this domain, so no modulus is needed. Split the integral: \[I=\int\frac{dx}{\sqrt{1-x}}-\int\sqrt{\frac{x}{1-x}}\,dx. \] The first part is $\displaystyle -2\sqrt{1-x}$. For the second put $\displaystyle x=\sin^{2}\varphi$, so $\displaystyle dx=2\sin\varphi\cos\varphi\,d\varphi$ and $\displaystyle \sqrt{\dfrac{x}{1-x}}=\tan\varphi$: \[\int\sqrt{\frac{x}{1-x}}\,dx=\int 2\sin^{2}\varphi\,d\varphi=\int(1-\cos 2\varphi)\,d\varphi=\varphi-\sin\varphi\cos\varphi=\sin^{-1}\sqrt{x}-\sqrt{x}\sqrt{1-x}. \] Therefore \[\int\sqrt{\frac{1-\sqrt{x}}{1+\sqrt{x}}}\,dx=-2\sqrt{1-x}-\sin^{-1}\sqrt{x}+\sqrt{x(1-x)}+\mathrm{C},\qquad 0\le x<1. \] (Equivalently $\displaystyle \cos^{-1}\sqrt{x}-2\sqrt{1-x}+\sqrt{x(1-x)}+\mathrm{C}$, which differs only by the constant $\displaystyle \frac{\pi}{2}$.)
Exercise 20
$\displaystyle \frac{2+\sin 2 x}{1+\cos 2 x} e^{x}$
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NCERT’s answer
$\displaystyle \mathrm{e}^{x} \tan x+\mathrm{C}$
Use the double-angle identities $\displaystyle 1+\cos 2x=2\cos^{2}x$ and $\displaystyle \sin 2x=2\sin x\cos x$: \[\frac{2+\sin 2x}{1+\cos 2x}=\frac{2+2\sin x\cos x}{2\cos^{2}x}=\frac{1}{\cos^{2}x}+\frac{\sin x}{\cos x}=\sec^{2}x+\tan x. \] The integral is now $\displaystyle \displaystyle\int e^{x}\left[\tan x+\sec^{2}x\right]dx$, which is exactly the form $\displaystyle \int e^{x}\left[f(x)+f'(x)\right]dx=e^{x}f(x)+\mathrm{C}$ with $\displaystyle f(x)=\tan x$, $\displaystyle f'(x)=\sec^{2}x$. Hence \[\int\frac{2+\sin 2x}{1+\cos 2x}\,e^{x}dx=e^{x}\tan x+\mathrm{C}. \]

Miscellaneous Exercise 21–30 (part 26 of 27)

Integrate the functions in Exercises $\displaystyle 1$ to 23.
Exercise 21
$\displaystyle \frac{x^{2}+x+1}{(x+1)^{2}(x+2)}$
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NCERT’s answer
$\displaystyle -2 \log |x+1|-\frac{1}{x+1}+3 \log |x+2|+C$
The factor $\displaystyle (x+1)$ is repeated, so it needs two partial-fraction terms: \[\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}=\frac{A}{x+1}+\frac{B}{(x+1)^{2}}+\frac{C}{x+2}, \] so $\displaystyle x^{2}+x+1=A(x+1)(x+2)+B(x+2)+C(x+1)^{2}$. Putting $\displaystyle x=-1$: $\displaystyle 1=B$. Putting $\displaystyle x=-2$: $\displaystyle 3=C$. Comparing coefficients of $\displaystyle x^{2}$: $\displaystyle 1=A+C$, so $\displaystyle A=-2$. \[\int\frac{x^{2}+x+1}{(x+1)^{2}(x+2)}dx=-2\int\frac{dx}{x+1}+\int\frac{dx}{(x+1)^{2}}+3\int\frac{dx}{x+2} \] \[=-2\log|x+1|-\frac{1}{x+1}+3\log|x+2|+\mathrm{C}=\log\left|\frac{(x+2)^{3}}{(x+1)^{2}}\right|-\frac{1}{x+1}+\mathrm{C}. \]
Exercise 22
$\displaystyle \tan ^{-1} \sqrt{\frac{1-x}{1+x}}$
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NCERT’s answer
$\displaystyle \frac{1}{2}\left[x \cos ^{-1} x-\sqrt{1-x^{2}}\right]+\mathrm{C}$
Simplify the inverse function first. Put $\displaystyle x=\cos 2\theta$, so $\displaystyle \theta=\frac12\cos^{-1}x$; then by $\displaystyle 1-\cos 2\theta=2\sin^{2}\theta$ and $\displaystyle 1+\cos 2\theta=2\cos^{2}\theta$, \[\tan^{-1}\sqrt{\frac{1-x}{1+x}}=\tan^{-1}\sqrt{\tan^{2}\theta}=\theta=\frac12\cos^{-1}x,\qquad -1<x\le 1, \] since $\displaystyle \theta\in\left[0,\frac{\pi}{2}\right)$ there, where $\displaystyle \tan\theta\ge 0$. So the integral is $\displaystyle \frac12\displaystyle\int\cos^{-1}x\,dx$. Integrating by parts with $\displaystyle \cos^{-1}x$ as the first function and $\displaystyle 1$ as the second, \[\int\cos^{-1}x\,dx=x\cos^{-1}x+\int\frac{x}{\sqrt{1-x^{2}}}dx=x\cos^{-1}x-\sqrt{1-x^{2}}. \] Hence \[\int\tan^{-1}\sqrt{\frac{1-x}{1+x}}\,dx=\frac12\left[x\cos^{-1}x-\sqrt{1-x^{2}}\right]+\mathrm{C}. \]
Exercise 23
$\displaystyle \frac{\sqrt{x^{2}+1}\left[\log \left(x^{2}+1\right)-2 \log x\right]}{x^{4}}$
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NCERT’s answer
$\displaystyle -\frac{1}{3}\left(1+\frac{1}{x^{2}}\right)^{\frac{3}{2}}\left[\log \left(1+\frac{1}{x^{2}}\right)-\frac{2}{3}\right]+\mathrm{C}$
Combine the logarithms: $\displaystyle \log\left(x^{2}+1\right)-2\log x=\log\dfrac{x^{2}+1}{x^{2}}=\log\left(1+\dfrac{1}{x^{2}}\right)$. Also, for $\displaystyle x>0$, $\displaystyle \dfrac{\sqrt{x^{2}+1}}{x^{4}}=\dfrac{x\sqrt{1+1/x^{2}}}{x^{4}}=\dfrac{1}{x^{3}}\sqrt{1+\dfrac{1}{x^{2}}}$. So the integrand is \[\frac{1}{x^{3}}\sqrt{1+\frac{1}{x^{2}}}\;\log\left(1+\frac{1}{x^{2}}\right). \] Put $\displaystyle t=1+\dfrac{1}{x^{2}}$, so $\displaystyle dt=-\dfrac{2}{x^{3}}dx$, i.e. $\displaystyle \dfrac{dx}{x^{3}}=-\dfrac{dt}{2}$: \[I=-\frac12\int t^{1/2}\log t\,dt. \] Integrate by parts with $\displaystyle \log t$ as the first function: \[\int t^{1/2}\log t\,dt=\frac23 t^{3/2}\log t-\frac23\int t^{1/2}dt=\frac23 t^{3/2}\log t-\frac49 t^{3/2}=\frac23 t^{3/2}\left(\log t-\frac23\right). \] Therefore \[I=-\frac13 t^{3/2}\left(\log t-\frac23\right)+\mathrm{C}=-\frac13\left(1+\frac{1}{x^{2}}\right)^{3/2}\left[\log\left(1+\frac{1}{x^{2}}\right)-\frac23\right]+\mathrm{C}. \]
Evaluate the definite integrals in Exercises $\displaystyle 24$ to 31.
Exercise 24
$\displaystyle \int_{\frac{\pi}{2}}^{\pi} e^{x}\left(\frac{1-\sin x}{1-\cos x}\right) d x$
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NCERT’s answer
$\displaystyle e^{\frac{\pi}{2}}$
Use the half-angle forms $\displaystyle 1-\cos x=2\sin^{2}\frac{x}{2}$ and $\displaystyle \sin x=2\sin\frac{x}{2}\cos\frac{x}{2}$: \[\frac{1-\sin x}{1-\cos x}=\frac{1-2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{2\sin^{2}\dfrac{x}{2}}=\frac12\,\mathrm{cosec}^{2}\frac{x}{2}-\cot\frac{x}{2}. \] Take $\displaystyle f(x)=-\cot\frac{x}{2}$; then $\displaystyle f'(x)=\frac12\mathrm{cosec}^{2}\frac{x}{2}$, so the integrand is $\displaystyle e^{x}\left[f(x)+f'(x)\right]$ and \[\int e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)dx=e^{x}f(x)+\mathrm{C}=-e^{x}\cot\frac{x}{2}+\mathrm{C}. \] Evaluating between the limits (note $\displaystyle \cot\frac{\pi}{2}=0$ and $\displaystyle \cot\frac{\pi}{4}=1$), \[\int_{\pi/2}^{\pi}e^{x}\left(\frac{1-\sin x}{1-\cos x}\right)dx=\left[-e^{x}\cot\frac{x}{2}\right]_{\pi/2}^{\pi}=-e^{\pi}\cdot 0+e^{\pi/2}\cdot 1=e^{\pi/2}. \]
Exercise 25
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x \cos x}{\cos ^{4} x+\sin ^{4} x} d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{8}$
Divide numerator and denominator by $\displaystyle \cos^{4}x$; this is legitimate on $\displaystyle \left[0,\frac{\pi}{4}\right]$, where $\displaystyle \cos x\neq 0$: \[\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}=\frac{\tan x\sec^{2}x}{1+\tan^{4}x}. \] Put $\displaystyle t=\tan^{2}x$, so $\displaystyle dt=2\tan x\sec^{2}x\,dx$. The limits change: $\displaystyle x=0\Rightarrow t=0$ and $\displaystyle x=\frac{\pi}{4}\Rightarrow t=1$. Hence \[\int_{0}^{\pi/4}\frac{\sin x\cos x}{\cos^{4}x+\sin^{4}x}dx=\frac12\int_{0}^{1}\frac{dt}{1+t^{2}}=\frac12\left[\tan^{-1}t\right]_{0}^{1}=\frac12\cdot\frac{\pi}{4}=\frac{\pi}{8}. \]
Exercise 26
$\displaystyle \int_{0}^{\frac{\pi}{2}} \frac{\cos ^{2} x d x}{\cos ^{2} x+4 \sin ^{2} x}$
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NCERT’s answer
$\displaystyle \frac{\pi}{6}$
Write the numerator in terms of the denominator. Using $\displaystyle 1=\cos^{2}x+\sin^{2}x$, look for constants $\displaystyle A,B$ with \[\cos^{2}x=A\left(\cos^{2}x+4\sin^{2}x\right)+B\left(\cos^{2}x+\sin^{2}x\right). \] Comparing coefficients of $\displaystyle \cos^{2}x$: $\displaystyle 1=A+B$; of $\displaystyle \sin^{2}x$: $\displaystyle 0=4A+B$. So $\displaystyle A=-\frac13$, $\displaystyle B=\frac43$, and \[\frac{\cos^{2}x}{\cos^{2}x+4\sin^{2}x}=-\frac13+\frac43\cdot\frac{1}{\cos^{2}x+4\sin^{2}x}=-\frac13+\frac43\cdot\frac{\sec^{2}x}{1+4\tan^{2}x}. \] For the second piece put $\displaystyle t=\tan x$, $\displaystyle dt=\sec^{2}x\,dx$; as $\displaystyle x\to\frac{\pi}{2}^{-}$ we have $\displaystyle t\to\infty$, so \[\frac43\int_{0}^{\infty}\frac{dt}{1+4t^{2}}=\frac43\cdot\frac12\left[\tan^{-1}2t\right]_{0}^{\infty}=\frac23\cdot\frac{\pi}{2}=\frac{\pi}{3}. \] Therefore \[\int_{0}^{\pi/2}\frac{\cos^{2}x\,dx}{\cos^{2}x+4\sin^{2}x}=-\frac13\cdot\frac{\pi}{2}+\frac{\pi}{3}=-\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{6}. \]
Exercise 27
$\displaystyle \int_{\frac{\pi}{6}}^{\frac{\pi}{3}} \frac{\sin x+\cos x}{\sqrt{\sin 2 x}} d x$
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NCERT’s answer
$\displaystyle 2 \sin ^{-1} \frac{(\sqrt{3}-1)}{2}$
The numerator $\displaystyle \sin x+\cos x$ is the derivative of $\displaystyle \sin x-\cos x$, and $\displaystyle \sin 2x$ can be expressed through that same quantity: \[(\sin x-\cos x)^{2}=1-2\sin x\cos x=1-\sin 2x\quad\Longrightarrow\quad \sin 2x=1-(\sin x-\cos x)^{2}. \] Put $\displaystyle t=\sin x-\cos x$, so $\displaystyle dt=(\cos x+\sin x)\,dx$. The limits become \[x=\frac{\pi}{6}\Rightarrow t=\frac12-\frac{\sqrt3}{2}=-\frac{\sqrt3-1}{2},\qquad x=\frac{\pi}{3}\Rightarrow t=\frac{\sqrt3}{2}-\frac12=\frac{\sqrt3-1}{2}. \] Hence \[\int_{\pi/6}^{\pi/3}\frac{\sin x+\cos x}{\sqrt{\sin 2x}}dx=\int_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}\frac{dt}{\sqrt{1-t^{2}}}=\left[\sin^{-1}t\right]_{-\frac{\sqrt3-1}{2}}^{\frac{\sqrt3-1}{2}}. \] Since $\displaystyle \sin^{-1}$ is an odd function, \[=2\sin^{-1}\left(\frac{\sqrt3-1}{2}\right). \]
Exercise 28
$\displaystyle \int_{0}^{1} \frac{d x}{\sqrt{1+x}-\sqrt{x}}$
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NCERT’s answer
$\displaystyle \frac{4 \sqrt{2}}{3}$
Rationalise the denominator; its conjugate reduces it to $\displaystyle (1+x)-x=1$: \[\frac{1}{\sqrt{1+x}-\sqrt{x}}=\frac{\sqrt{1+x}+\sqrt{x}}{(1+x)-x}=\sqrt{1+x}+\sqrt{x}. \] Hence \[\int_{0}^{1}\frac{dx}{\sqrt{1+x}-\sqrt{x}}=\int_{0}^{1}\left(\sqrt{1+x}+\sqrt{x}\right)dx=\left[\frac23 (1+x)^{3/2}+\frac23 x^{3/2}\right]_{0}^{1} \] \[=\left(\frac23\cdot 2\sqrt2+\frac23\right)-\left(\frac23+0\right)=\frac{4\sqrt2}{3}. \]
Exercise 29
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$
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NCERT’s answer
$\displaystyle \frac{1}{40} \log 9$
Again $\displaystyle \sin x+\cos x$ is the derivative of $\displaystyle \sin x-\cos x$. Put $\displaystyle t=\sin x-\cos x$, so $\displaystyle dt=(\sin x+\cos x)\,dx$ and $\displaystyle \sin 2x=1-t^{2}$. The limits become $\displaystyle x=0\Rightarrow t=-1$ and $\displaystyle x=\frac{\pi}{4}\Rightarrow t=0$, and the denominator becomes $\displaystyle 9+16\left(1-t^{2}\right)=25-16t^{2}$: \[I=\int_{-1}^{0}\frac{dt}{25-16t^{2}}=\frac{1}{16}\int_{-1}^{0}\frac{dt}{\left(\frac54\right)^{2}-t^{2}}. \] Using $\displaystyle \int\frac{dt}{a^{2}-t^{2}}=\frac{1}{2a}\log\left|\frac{a+t}{a-t}\right|$ with $\displaystyle a=\frac54$, \[I=\frac{1}{16}\cdot\frac{2}{5}\left[\log\left|\frac{5+4t}{5-4t}\right|\right]_{-1}^{0}=\frac{1}{40}\left(\log 1-\log\frac19\right)=\frac{1}{40}\log 9=\frac{1}{20}\log 3. \]
Exercise 30
$\displaystyle \int_{0}^{\frac{\pi}{2}} \sin 2 x \tan ^{-1}(\sin x) d x$
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NCERT’s answer
$\displaystyle \frac{\pi}{2}-1$
Put $\displaystyle t=\sin x$, so $\displaystyle dt=\cos x\,dx$ and $\displaystyle \sin 2x\,dx=2\sin x\cos x\,dx=2t\,dt$; the limits become $\displaystyle 0$ and $\displaystyle 1$: \[I=\int_{0}^{\pi/2}\sin 2x\,\tan^{-1}(\sin x)\,dx=\int_{0}^{1}2t\tan^{-1}t\,dt. \] Integrate by parts with $\displaystyle \tan^{-1}t$ as the first function and $\displaystyle 2t$ as the second: \[I=\left[t^{2}\tan^{-1}t\right]_{0}^{1}-\int_{0}^{1}\frac{t^{2}}{1+t^{2}}dt=\frac{\pi}{4}-\int_{0}^{1}\left(1-\frac{1}{1+t^{2}}\right)dt, \] where $\displaystyle \dfrac{t^{2}}{1+t^{2}}$ was made proper before integrating. Hence \[I=\frac{\pi}{4}-\left[t-\tan^{-1}t\right]_{0}^{1}=\frac{\pi}{4}-\left(1-\frac{\pi}{4}\right)=\frac{\pi}{2}-1. \]

Miscellaneous Exercise 31–40 (part 27 of 27)

Evaluate the definite integrals in Exercises $\displaystyle 24$ to 31.
Exercise 31
$\displaystyle \int_{1}^{4}[|x-1|+|x-2|+|x-3|] d x$
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NCERT’s answer
$\displaystyle \frac{19}{2}$
A modulus must be split at the point where its argument changes sign, and $\displaystyle x=2$ and $\displaystyle x=3$ both lie inside $\displaystyle [1,4]$. On $\displaystyle [1,4]$ the first modulus needs no split, since $\displaystyle x-1\ge 0$ throughout. \[\int_{1}^{4}|x-1|\,dx=\int_{1}^{4}(x-1)dx=\left[\frac{(x-1)^{2}}{2}\right]_{1}^{4}=\frac92. \] \[\int_{1}^{4}|x-2|\,dx=\int_{1}^{2}(2-x)dx+\int_{2}^{4}(x-2)dx=\frac12+2=\frac52. \] \[\int_{1}^{4}|x-3|\,dx=\int_{1}^{3}(3-x)dx+\int_{3}^{4}(x-3)dx=2+\frac12=\frac52. \] Adding the three, \[\int_{1}^{4}\big[|x-1|+|x-2|+|x-3|\big]dx=\frac92+\frac52+\frac52=\frac{19}{2}. \]
Prove the following (Exercises $\displaystyle 32$ to $\displaystyle 37$)
Exercise 32
$\displaystyle \int_{1}^{3} \frac{d x}{x^{2}(x+1)}=\frac{2}{3}+\log \frac{2}{3}$
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The denominator carries the repeated factor $\displaystyle x^{2}$, so the partial-fraction form is \[\frac{1}{x^{2}(x+1)}=\frac{A}{x}+\frac{B}{x^{2}}+\frac{C}{x+1},\qquad 1=Ax(x+1)+B(x+1)+Cx^{2}. \] Putting $\displaystyle x=0$: $\displaystyle B=1$. Putting $\displaystyle x=-1$: $\displaystyle C=1$. Comparing coefficients of $\displaystyle x^{2}$: $\displaystyle 0=A+C$, so $\displaystyle A=-1$. Hence \[\int\frac{dx}{x^{2}(x+1)}=-\log|x|-\frac1x+\log|x+1|=\log\left|\frac{x+1}{x}\right|-\frac1x. \] Both limits are positive, so the moduli may be dropped: \[\int_{1}^{3}\frac{dx}{x^{2}(x+1)}=\left(\log\frac43-\frac13\right)-\left(\log 2-1\right)=\log\frac{4}{3\cdot 2}+\left(1-\frac13\right)=\frac23+\log\frac23. \] This is the required result, $\displaystyle \displaystyle\int_{1}^{3}\frac{dx}{x^{2}(x+1)}=\frac23+\log\frac23$.
Exercise 33
$\displaystyle \int_{0}^{1} x e^{x} d x=1$
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Integrate by parts, taking $\displaystyle x$ as the first function and $\displaystyle e^{x}$ as the second (ILATE: algebraic before exponential): \[\int_{0}^{1}xe^{x}dx=\left[xe^{x}\right]_{0}^{1}-\int_{0}^{1}1\cdot e^{x}dx=\left[xe^{x}-e^{x}\right]_{0}^{1}. \] Evaluating, \[=\left(1\cdot e-e\right)-\left(0\cdot 1-1\right)=0-(-1)=1. \] Hence $\displaystyle \displaystyle\int_{0}^{1}xe^{x}dx=1$, as required.
Exercise 34
$\displaystyle \int_{-1}^{1} x^{17} \cos ^{4} x d x=0$
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This solution has not been cross-checked against the answer printed in NCERT.
Use the property that $\displaystyle \int_{-a}^{a}f(x)dx=0$ whenever $\displaystyle f$ is odd, i.e. $\displaystyle f(-x)=-f(x)$. Let $\displaystyle f(x)=x^{17}\cos^{4}x$. Then, since $\displaystyle 17$ is odd and $\displaystyle \cos$ is an even function, \[f(-x)=(-x)^{17}\cos^{4}(-x)=-x^{17}\cos^{4}x=-f(x). \] So $\displaystyle f$ is odd, and the interval $\displaystyle [-1,1]$ is symmetric about the origin. Therefore \[\int_{-1}^{1}x^{17}\cos^{4}x\,dx=0. \]
Exercise 35
$\displaystyle \int_{0}^{\frac{\pi}{2}} \sin ^{3} x d x=\frac{2}{3}$
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This solution has not been cross-checked against the answer printed in NCERT.
Split off one factor of $\displaystyle \sin x$ and convert the rest, using $\displaystyle \sin^{2}x=1-\cos^{2}x$: \[\sin^{3}x=\sin x\left(1-\cos^{2}x\right). \] Put $\displaystyle t=\cos x$, so $\displaystyle dt=-\sin x\,dx$; the limits become $\displaystyle x=0\Rightarrow t=1$ and $\displaystyle x=\frac{\pi}{2}\Rightarrow t=0$ (note the reversal, which the minus sign cancels): \[\int_{0}^{\pi/2}\sin^{3}x\,dx=-\int_{1}^{0}\left(1-t^{2}\right)dt=\int_{0}^{1}\left(1-t^{2}\right)dt=\left[t-\frac{t^{3}}{3}\right]_{0}^{1}=1-\frac13=\frac23. \] Hence $\displaystyle \displaystyle\int_{0}^{\pi/2}\sin^{3}x\,dx=\frac23$, as required.
Exercise 36
$\displaystyle \int_{0}^{\frac{\pi}{4}} 2 \tan ^{3} x d x=1-\log 2$
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This solution has not been cross-checked against the answer printed in NCERT.
Reduce the power using $\displaystyle \tan^{2}x=\sec^{2}x-1$: \[\tan^{3}x=\tan x\left(\sec^{2}x-1\right)=\tan x\sec^{2}x-\tan x. \] With $\displaystyle \int\tan x\sec^{2}x\,dx=\frac{\tan^{2}x}{2}$ (put $\displaystyle u=\tan x$) and $\displaystyle \int\tan x\,dx=-\log|\cos x|$, \[\int 2\tan^{3}x\,dx=2\left[\frac{\tan^{2}x}{2}+\log|\cos x|\right]=\tan^{2}x+2\log|\cos x|. \] Evaluating between $\displaystyle 0$ and $\displaystyle \frac{\pi}{4}$, where $\displaystyle \tan\frac{\pi}{4}=1$ and $\displaystyle \cos\frac{\pi}{4}=\frac{1}{\sqrt2}$, \[\int_{0}^{\pi/4}2\tan^{3}x\,dx=\left(1+2\log\frac{1}{\sqrt2}\right)-\left(0+2\log 1\right)=1-\log 2. \] Hence $\displaystyle \displaystyle\int_{0}^{\pi/4}2\tan^{3}x\,dx=1-\log 2$, as required.
Exercise 37
$\displaystyle \int_{0}^{1} \sin ^{-1} x d x=\frac{\pi}{2}-1$
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This solution has not been cross-checked against the answer printed in NCERT.
Integrate by parts, taking $\displaystyle \sin^{-1}x$ as the first function and $\displaystyle 1$ as the second: \[\int_{0}^{1}\sin^{-1}x\,dx=\left[x\sin^{-1}x\right]_{0}^{1}-\int_{0}^{1}\frac{x}{\sqrt{1-x^{2}}}dx. \] The first bracket is $\displaystyle 1\cdot\frac{\pi}{2}-0=\frac{\pi}{2}$. For the remaining integral put $\displaystyle u=1-x^{2}$, $\displaystyle du=-2x\,dx$, giving \[\int_{0}^{1}\frac{x\,dx}{\sqrt{1-x^{2}}}=\left[-\sqrt{1-x^{2}}\right]_{0}^{1}=0-(-1)=1. \] Hence \[\int_{0}^{1}\sin^{-1}x\,dx=\frac{\pi}{2}-1, \] as required.
Choose the correct answers in Exercises $\displaystyle 38$ to $\displaystyle 40$
Exercise 38
$\displaystyle \int \frac{d x}{e^{x}+e^{-x}}$ is equal to (A) $\displaystyle \tan ^{-1}\left(e^{x}\right)+\mathrm{C}$ (B) $\displaystyle \tan ^{-1}\left(e^{-x}\right)+\mathrm{C}$ (C) $\displaystyle \log \left(e^{x}-e^{-x}\right)+\mathrm{C}$ (D) $\displaystyle \log \left(e^{x}+e^{-x}\right)+\mathrm{C}$
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
A
Multiply numerator and denominator by $\displaystyle e^{x}$, so that the derivative needed for the substitution appears: \[\int\frac{dx}{e^{x}+e^{-x}}=\int\frac{e^{x}\,dx}{e^{2x}+1}. \] Put $\displaystyle t=e^{x}$, so $\displaystyle dt=e^{x}dx$ and $\displaystyle e^{2x}=t^{2}$: \[=\int\frac{dt}{1+t^{2}}=\tan^{-1}t+\mathrm{C}=\tan^{-1}\left(e^{x}\right)+\mathrm{C}. \] So the correct answer is (A).
Exercise 39
$\displaystyle \int \frac{\cos 2 x}{(\sin x+\cos x)^{2}} d x$ is equal to (A) $\displaystyle \frac{-1}{\sin x+\cos x}+C$ (B) $\displaystyle \log |\sin x+\cos x|+C$ (C) $\displaystyle \log |\sin x-\cos x|+\mathrm{C}$ (D) $\displaystyle \frac{1}{(\sin x+\cos x)^{2}}$
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
B
Factorise the numerator, $\displaystyle \cos 2x=\cos^{2}x-\sin^{2}x=(\cos x-\sin x)(\cos x+\sin x)$, and cancel one factor of $\displaystyle (\sin x+\cos x)$: \[\int\frac{\cos 2x}{(\sin x+\cos x)^{2}}dx=\int\frac{\cos x-\sin x}{\sin x+\cos x}dx. \] The numerator is now exactly the derivative of the denominator, so put $\displaystyle t=\sin x+\cos x$, $\displaystyle dt=(\cos x-\sin x)dx$: \[=\int\frac{dt}{t}=\log|t|+\mathrm{C}=\log|\sin x+\cos x|+\mathrm{C}. \] So the correct answer is (B).
Exercise 40
If $\displaystyle f(a+b-x)=f(x)$, then $\displaystyle \int_{a}^{b} x f(x) d x$ is equal to (A) $\displaystyle \frac{a+b}{2} \int_{a}^{b} f(b-x) d x$ (B) $\displaystyle \frac{a+b}{2} \int_{a}^{b} f(b+x) d x$ (C) $\displaystyle \frac{b-a}{2} \int_{a}^{b} f(x) d x$ (D) $\displaystyle \frac{a+b}{2} \int_{a}^{b} f(x) d x$
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This solution has not been cross-checked against the answer printed in NCERT.
NCERT’s answer
D
Use the property $\displaystyle \displaystyle\int_{a}^{b}g(x)\,dx=\int_{a}^{b}g(a+b-x)\,dx$. Writing $\displaystyle I$ for the required integral, \[I=\int_{a}^{b}xf(x)\,dx=\int_{a}^{b}(a+b-x)\,f(a+b-x)\,dx. \] The hypothesis $\displaystyle f(a+b-x)=f(x)$ now replaces the second factor: \[I=\int_{a}^{b}(a+b-x)f(x)\,dx=(a+b)\int_{a}^{b}f(x)\,dx-\int_{a}^{b}xf(x)\,dx=(a+b)\int_{a}^{b}f(x)\,dx-I. \] Bringing $\displaystyle I$ to the left, $\displaystyle 2I=(a+b)\displaystyle\int_{a}^{b}f(x)\,dx$, so \[\int_{a}^{b}xf(x)\,dx=\frac{a+b}{2}\int_{a}^{b}f(x)\,dx. \] So the correct answer is (D).

EXERCISE 7.1 1–10 (part 1 of 27)

Exercise 1

Exercise 2

Exercise 3

Exercise 4

Exercise 5

Exercise 6

Exercise 7

Exercise 8

Exercise 9

Exercise 10