NCERT Solutions · Class 12 Mathematics Integrals

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 \[\int\frac{\sin x}{(1+\cos x)^{2}}\,dx=\frac{1}{1+\cos x}+\mathrm{C},\qquad \cos x\neq-1.\]

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 \[\int\frac{1}{1+\cot x}\,dx=\frac{x}{2}-\frac{1}{2}\log\left|\sin x+\cos x\right|+\mathrm{C}.\]

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 \[\int\frac{1}{1-\tan x}\,dx=\frac{x}{2}-\frac{1}{2}\log\left|\cos x-\sin x\right|+\mathrm{C}.\]

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 \(\displaystyle \tan x>0\).

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 \[\int\frac{(1+\log x)^{2}}{x}\,dx=\frac{(1+\log x)^{3}}{3}+\mathrm{C},\qquad x>0.\]

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 \[\int\frac{(x+1)(x+\log x)^{2}}{x}\,dx=\frac{(x+\log x)^{3}}{3}+\mathrm{C},\qquad x>0.\]

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 \[\int\frac{x^{3}\sin\left(\tan^{-1}x^{4}\right)}{1+x^{8}}\,dx=-\frac{1}{4}\cos\left(\tan^{-1}x^{4}\right)+\mathrm{C}.\]

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) \(\displaystyle \log\left(10^{x}+x^{10}\right)+\mathrm{C}\).

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) \(\displaystyle \tan x-\cot x+\mathrm{C}\).