SolveIt is under development
SolveItNCERT · CBSE · NEET

NCERT Exemplar · Class 12 Mathematics Integrals

63 questions · 63 still being checked

EXERCISE 7.3 21–30 (part 3 of 6)

  1. Evaluate the following:

    Exercise 21

    ∫sin⁡−1x(1−x2)32dx\displaystyle \int \frac{\sin ^{-1} x}{\left(1-x^2\right)^{\frac{3}{2}}} d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{x \sin ^{-1} x}{\sqrt{1-x^2}}+\log \left|\sqrt{1-x^2}\right|\)
    Put \(\displaystyle x=\sin\theta\), so \(\displaystyle dx=\cos\theta\,d\theta\) and \(\displaystyle (1-x^2)^{3/2}=\cos^3\theta\). \[\int\frac{\sin^{-1}x}{(1-x^2)^{3/2}}\,dx=\int\frac{\theta\cos\theta}{\cos^3\theta}\,d\theta=\int\theta\sec^2\theta\,d\theta \] By parts: \[\int\theta\sec^2\theta\,d\theta=\theta\tan\theta-\int\tan\theta\,d\theta=\theta\tan\theta+\ln|\cos\theta|+C \] \[\tan\theta=\frac{x}{\sqrt{1-x^2}},\qquad \cos\theta=\sqrt{1-x^2} \] \[\int\frac{\sin^{-1}x}{(1-x^2)^{3/2}}\,dx=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}+\frac12\ln(1-x^2)+C \] Answer: \(\displaystyle \dfrac{x\sin^{-1}x}{\sqrt{1-x^2}}+\dfrac12\ln(1-x^2)+C\)

  2. Exercise 22

    ∫(cos⁡5x+cos⁡4x)1−2cos⁡3xdx\displaystyle \int \frac{(\cos 5 x+\cos 4 x)}{1-2 \cos 3 x} d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle -\left(\frac{1}{2} \sin 2 x+\sin x\right)+c\)
    Product to sum: \[2\cos3x\cos2x=\cos5x+\cos x,\qquad 2\cos3x\cos x=\cos4x+\cos2x \] \[\cos5x+\cos4x=2\cos3x\,(\cos2x+\cos x)-(\cos x+\cos2x) \] \[\cos5x+\cos4x=-(1-2\cos3x)(\cos2x+\cos x) \] \[\frac{\cos5x+\cos4x}{1-2\cos3x}=-(\cos2x+\cos x) \] \[\int-(\cos2x+\cos x)\,dx=-\frac{\sin2x}{2}-\sin x+C \] Answer: \(\displaystyle -\dfrac{\sin2x}{2}-\sin x+C\)

  3. Exercise 23

    ∫sin⁡6x+cos⁡6xsin⁡2xcos⁡2xdx\displaystyle \int \frac{\sin ^6 x+\cos ^6 x}{\sin ^2 x \cos ^2 x} d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \tan x-\cot x-3 x+c\)
    \[\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x\,(\sin^2x+\cos^2x)=1-3\sin^2x\cos^2x \] \[\frac{\sin^6x+\cos^6x}{\sin^2x\cos^2x}=\frac{\sin^2x+\cos^2x}{\sin^2x\cos^2x}-3=\sec^2x+\operatorname{cosec}^2x-3 \] \[\int(\sec^2x+\operatorname{cosec}^2x-3)\,dx=\tan x-\cot x-3x+C \] Answer: \(\displaystyle \tan x-\cot x-3x+C\)

  4. Exercise 24

    ∫xa3−x3dx\displaystyle \int \frac{\sqrt{x}}{\sqrt{a^3-x^3}} d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{2}{3} \sin ^{-1} \sqrt{\frac{x^3}{a^3}}+c\)
    Put \(\displaystyle x^{3/2}=t\), so \(\displaystyle x^3=t^2\) and \(\displaystyle \tfrac32\sqrt{x}\,dx=dt\). \[\int\frac{\sqrt{x}\,dx}{\sqrt{a^3-x^3}}=\frac23\int\frac{dt}{\sqrt{(a^{3/2})^2-t^2}} \] \[=\frac23\sin^{-1}\frac{t}{a^{3/2}}+C \] \[=\frac23\sin^{-1}\left[\left(\frac{x}{a}\right)^{3/2}\right]+C \] Answer: \(\displaystyle \dfrac23\sin^{-1}\left[\left(\dfrac{x}{a}\right)^{3/2}\right]+C\)

  5. Exercise 25

    ∫cos⁡x−cos⁡2x1−cos⁡xdx\displaystyle \int \frac{\cos x-\cos 2 x}{1-\cos x} d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle 2 \sin x+x+c\)
    Factorise the numerator.\[\cos x - \cos 2x = \cos x + 1 - 2\cos^2 x = (1-\cos x)(1+2\cos x) \]\[\int \frac{\cos x-\cos 2x}{1-\cos x}\,dx = \int (1+2\cos x)\,dx \]\[= x + 2\sin x + C \]Answer: \(\displaystyle x + 2\sin x + C\)

  6. Exercise 26

    ∫dxxx4−1\displaystyle \int \frac{d x}{x \sqrt{x^4-1}} (Hint : Put x2=sec⁡θ\displaystyle x^2=\sec \theta)

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{1}{2} \sec ^{-1}\left(x^2\right)+c\)
    Put \(\displaystyle x^2=\sec\theta\).\[2x\,dx = \sec\theta\tan\theta\,d\theta \]\[\sqrt{x^4-1} = \sqrt{\sec^2\theta-1} = \tan\theta \]\[\frac{dx}{x\sqrt{x^4-1}} = \frac{x\,dx}{x^2\sqrt{x^4-1}} = \frac{\tfrac12\sec\theta\tan\theta\,d\theta}{\sec\theta\,\tan\theta} = \frac{d\theta}{2} \]\[\int\frac{d\theta}{2} = \frac{\theta}{2}+C = \frac12\sec^{-1}\!\left(x^2\right)+C \]Answer: \(\displaystyle \dfrac12\sec^{-1}\!\left(x^2\right)+C\)
  7. Evaluate the following as limit of sums:

    Exercise 27

    ∫02(x2+3)dx\displaystyle \int_0^2\left(x^2+3\right) d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{26}{3}\)
    Here \(\displaystyle a=0,\ b=2,\ f(x)=x^2+3,\ h=\dfrac2n\), so \(\displaystyle nh=2\).\[\int_0^2 f(x)\,dx = \lim_{n\to\infty} h\sum_{r=0}^{n-1} f(rh) \]\[= \lim_{n\to\infty} h\sum_{r=0}^{n-1}\left(r^2h^2+3\right) \]\[= \lim_{n\to\infty}\left[h^3\cdot\frac{(n-1)n(2n-1)}{6} + 3nh\right] \]\[= \lim_{n\to\infty}\left[\frac{8}{n^3}\cdot\frac{(n-1)n(2n-1)}{6} + 6\right] \]\[= \lim_{n\to\infty}\left[\frac43\left(1-\frac1n\right)\left(2-\frac1n\right)+6\right] \]\[= \frac83 + 6 = \frac{26}{3} \]Answer: \(\displaystyle \dfrac{26}{3}\)

  8. Exercise 28

    ∫02exdx\displaystyle \int_0^2 e^x d x

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle e^2-1\)
    Here \(\displaystyle a=0,\ b=2,\ f(x)=e^x,\ h=\dfrac2n\), so \(\displaystyle nh=2\).\[\int_0^2 e^x\,dx = \lim_{h\to0} h\left[1+e^h+e^{2h}+\dots+e^{(n-1)h}\right] \]The bracket is a G.P. with ratio \(\displaystyle e^h\).\[= \lim_{h\to0} h\cdot\frac{e^{nh}-1}{e^h-1} \]\[= \left(e^2-1\right)\lim_{h\to0}\frac{h}{e^h-1} \]\[\lim_{h\to0}\frac{e^h-1}{h}=1 \]\[= e^2-1 \]Answer: \(\displaystyle e^2-1\)
  9. Evaluate the following:

    Exercise 29

    ∫01dxex+e−x\displaystyle \int_0^1 \frac{d x}{e^x+e^{-x}}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \tan ^{-1} e-\frac{\pi}{4}\)
    \[\frac{1}{e^x+e^{-x}} = \frac{e^x}{e^{2x}+1} \]Put \(\displaystyle t=e^x\), so \(\displaystyle dt=e^x\,dx\); \(\displaystyle x:0\to1\) gives \(\displaystyle t:1\to e\).\[I = \int_1^e \frac{dt}{1+t^2} \]\[= \left[\tan^{-1}t\right]_1^e = \tan^{-1}e - \tan^{-1}1 \]\[= \tan^{-1}e - \frac{\pi}{4} \]Answer: \(\displaystyle \tan^{-1}e-\dfrac{\pi}{4}\)

  10. Exercise 30

    ∫0π2tan⁡xdx1+m2tan⁡2x\displaystyle \int_0^{\frac{\pi}{2}} \frac{\tan x d x}{1+m^2 \tan ^2 x}

    Matches the book, not yet reviewed

    This working reaches the answer NCERT prints. It has not yet been read through by hand.

    NCERT’s answer
    \(\displaystyle \frac{\log m}{m^2-1}\)
    Multiply numerator and denominator by \(\displaystyle \cos^2x\).\[\frac{\tan x}{1+m^2\tan^2x} = \frac{\sin x\cos x}{\cos^2x+m^2\sin^2x} \]Put \(\displaystyle t=\sin^2x\), so \(\displaystyle dt=2\sin x\cos x\,dx\); \(\displaystyle x:0\to\frac\pi2\) gives \(\displaystyle t:0\to1\).\[\cos^2x+m^2\sin^2x = 1+(m^2-1)t \]\[I = \frac12\int_0^1\frac{dt}{1+(m^2-1)t} \]For \(\displaystyle m^2\ne1\):\[I = \frac{1}{2(m^2-1)}\Big[\ln\big(1+(m^2-1)t\big)\Big]_0^1 = \frac{\ln m^2}{2(m^2-1)} \]\[I = \frac{\ln|m|}{m^2-1} \]For \(\displaystyle m^2=1\):\[I = \frac12\int_0^1dt = \frac12 \]\(\displaystyle m=0\) is excluded: \(\displaystyle \int_0^{\pi/2}\tan x\,dx\) diverges.Answer: \(\displaystyle \dfrac{\ln|m|}{m^2-1}\) for \(\displaystyle m\ne0,\pm1\); \(\displaystyle \dfrac12\) for \(\displaystyle m=\pm1\)