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
xsec2x
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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
tan−1x
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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
x(logx)2
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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
(x2+1)logx
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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
ex(sinx+cosx)
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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
(1+x)2xex
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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
ex(1+cosx1+sinx)
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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
ex(x1−x21)
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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
(x−1)3(x−3)ex
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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 \(\displaystyle x\neq 1\).