JEE Main 2019 — Definite Integration Question with Solution

Given integral, $I={\int }_0^{1}x{\cot }^{-1}\left(1-x^2+x^4\right)dx$

Put $x^2=t, \Rightarrow 2xdx=dt$

$\Rightarrow I=\frac{1}{2}{\int }_0^1{\cot }^{-1}\left(1-t+t^2\right)dt$

$\Rightarrow I=\frac{1}{2}{\int }_0^1{\tan }^{-1}\left(\frac{1}{1-t+t^2}\right)dt$

$\Rightarrow I=\frac{1}{2}{\int }_0^1{\tan }^{-1}\left(\frac{t-\left(t-1\right)}{1+t\left(t-1\right)}\right)dt$

Using ${\tan }^{-1}a-{\tan }^{-1}b={\tan }^{-1}\left(\frac{a-b}{1+a·b}\right),$ we get

$\Rightarrow I=\frac{1}{2}\left[{\int }_0^1\left({\tan }^{-1}t\right)dt-{\int }_0^1{\tan }^{-1}\left(t-1\right)dt\right]$

Using ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx,$ we get

$\Rightarrow I=\frac{1}{2}\left[{\int }_0^1\left({\tan }^{-1}t\right)dt-{\int }_0^1{\tan }^{-1}\left(\left(1-t\right)-1\right)dt\right]$

$\Rightarrow I=\frac{1}{2}\left[{\int }_0^1\left({\tan }^{-1}t\right)dt-{\int }_0^1{\tan }^{-1}\left(-t\right)dt\right]$

Now, using ${\tan }^{-1}\left(-a\right)=-{\tan }^{-1}a,$ we get

$\Rightarrow I=\frac{1}{2}\left[{\int }_0^1\left({\tan }^{-1}t\right)dt+{\int }_0^1\left({\tan }^{-1}t\right)dt\right]$

$\Rightarrow I=\frac{1}{2}\left[2{\int }_0^1{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}tdt\right]$

$\Rightarrow I={\int }_0^1\left(1\cdot {\tan }^{-1}t\right)dt$

Now applying the integration by parts i.e. ${\int }_a^b\left(u·v\right)dx=u{\int }_a^{b}vdx-{\int }_a^b\left[\frac{d}{dx}\left(u\right)\int vdx\right]dx+c$

$\Rightarrow I={{\mathrm{t}\mathrm{a}\mathrm{n}}^{-1}t\cdot t|}_0^1-{\int }_0^1\left(\frac{1}{1+t^2}\cdot t\right)dt$

Put $1+t^2=u, \Rightarrow 2tdt=du$

$\Rightarrow I={\tan }^{-1}\left(1\right)\cdot 1-0-\frac{1}{2}{\int }_1^2\left(\frac{1}{u}\right)du$

$\Rightarrow I=\frac{\pi }{4}-\frac{1}{2}{\log }_2\left(u\right){\left.\mathrm{}\right|}_1^2$

$\Rightarrow I=\frac{\pi }{4}-\frac{1}{2}{\log }_{e}2.$