Question

Evaluate the definite integral ${\int }_0^{\frac{\pi }{2}}\sin 2x{\tan }^{-1}\left(\sin x\right)dx$

Answer 1

Let $I={\int }_0^{\frac{\pi }{2}}\sin 2x{\tan }^{-1}\left(\sin x\right)dx={\int }_0^{\frac{\pi }{2}}2\sin x\cos x{\tan }^{-1}\left(\sin x\right)dx$
Also, let $\sin x=t\Rightarrow \cos xdx=dt$
When $x=0,t=0$ and when $x=\frac{\pi }{2},t=1$
$\Rightarrow I=2{\int }_0^1{\tan }^{-1}\left(t\right)dt$ ............ (1)
Consider $\int t\cdot {\tan }^{-1}tdt={\tan }^{-1}t\cdot \int tdt-\int \left\{\frac{d}{dt}\right({\tan }^{-1}t)\int tdt\}dt$
$={\tan }^{-1}t\cdot \frac{t^2}{2}-\int \frac{1}{1+t^2}\cdot \frac{t^2}{2}dt$
$=\frac{t^2{\tan }^{-1}t}{2}-\frac{1}{2}\int \frac{t^2+1-1}{1+t^2}dt$
$=\frac{t^2{\tan }^{-1}t}{2}-\frac{1}{2}\int 1dt+\frac{1}{2}\int \frac{1}{1+t^2}dt$
$=\frac{t^2{\tan }^{-1}t}{2}-\frac{1}{2}\cdot t+\frac{1}{2}{\tan }^{-1}t$
$\Rightarrow {\int }_0^{1}t\cdot {\tan }^{-1}tdt={[\frac{t^2\cdot {\tan }^{-1}t}{2}-\frac{t}{2}+\frac{1}{2}{\tan }^{-1}t]}_0^1$
$=\frac{1}{2}[\frac{\pi }{4}-1+\frac{\pi }{4}]$
$=\frac{1}{2}[\frac{\pi }{2}-1]=\frac{\pi }{4}-\frac{1}{2}$
From equation (1), we obtain
$I=2[\frac{\pi }{4}-\frac{1}{2}]=\frac{\pi }{2}-1$