Question

Show that ${\int }_0^{\pi /2}{x}^2\sin 2x.{\tan }^{-1}\left(\sin x\right)dx$

Answer 1

Let $I={\int }_0^{\pi /2}{x}^2\sin 2x.{\tan }^{-1}\left(\sin x\right)dx$

$={\int }_0^{\pi /2}2.\sin x\cos x{\tan }^{-1}\left(\sin x\right)dx$

Put $\sin x=\theta$

$\Rightarrow \cos xdx=d\theta$

When $x=0\Rightarrow \theta =0$

When $x=\frac{\pi }{2}\Rightarrow \theta =1$

$\therefore I=2{\int }_0^1\theta .{\tan }^{-1}\theta d\theta$

Applying integration by parts, we get

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

$=\frac{\pi }{4}-{\int }_0^1(1-\frac{1}{1+{\theta }^2})d\theta$

$\Rightarrow I=\frac{\pi }{4}-{[\theta -{\tan }^{-1}\theta ]}_0^1=\frac{\pi }{2}-1$