NCERT Solutions · Class 12 Mathematics Integrals

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 \(\displaystyle \int\tan u\,du=-\log|\cos u|\), \[\int\frac{dx}{\cos (x+a)\cos (x+b)}=\frac{1}{\sin (a-b)}\Big[-\log|\cos (x+a)|+\log|\cos (x+b)|\Big]+\mathrm{C} \] \[=\frac{1}{\sin (a-b)}\log\left|\frac{\cos (x+b)}{\cos (x+a)}\right|+\mathrm{C}. \]

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 \(\displaystyle dt=4x^{3}dx\): \[\int\frac{x^{3}}{\sqrt{1-x^{8}}}dx=\frac14\int\frac{dt}{\sqrt{1-t^{2}}}=\frac14\sin^{-1}t+\mathrm{C}=\frac14\sin^{-1}\left(x^{4}\right)+\mathrm{C}. \]

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.

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 \(\displaystyle \int\frac{dx}{x^{2}+a^{2}}=\frac1a\tan^{-1}\frac{x}{a}\), \[\int\frac{dx}{\left(x^{2}+1\right)\left(x^{2}+4\right)}=\frac13\left[\tan^{-1}x-\frac12\tan^{-1}\frac{x}{2}\right]+\mathrm{C}=\frac13\tan^{-1}x-\frac16\tan^{-1}\frac{x}{2}+\mathrm{C}. \]

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 \(\displaystyle dt=-\sin x\,dx\): \[\int\cos^{3}x\,\sin x\,dx=-\int t^{3}dt=-\frac{t^{4}}{4}+\mathrm{C}=-\frac{\cos^{4}x}{4}+\mathrm{C}. \]

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 \(\displaystyle x^{4}+1>0\).

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 \(\displaystyle \frac1a\log\big|f(ax+b)\big|+\mathrm{C}\).)

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 \(\displaystyle \cos\alpha+\sin\alpha\cot x=\dfrac{\sin x\cos\alpha+\cos x\sin\alpha}{\sin x}=\dfrac{\sin (x+\alpha)}{\sin x}\), \[=-\frac{2}{\sin\alpha}\sqrt{\frac{\sin (x+\alpha)}{\sin x}}+\mathrm{C},\qquad \sin\alpha\neq 0. \]

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 \(\displaystyle \frac{\pi}{2}\).)

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 \[\int\frac{2+\sin 2x}{1+\cos 2x}\,e^{x}dx=e^{x}\tan x+\mathrm{C}. \]