Question

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

Answer 1

${\int }_0^{\frac{\pi }{2}}\sin 2xta{n}^{-1}\left(\sin x\right)dx$

$={\int }_0^{\frac{\pi }{2}}2\sin x\cos xta{n}^{-1}\left(\sin x\right)dx$

substitute $u=\sin x\to du=\cos xdx$

$=2{\int }_0^{\frac{\pi }{2}}\left(uta{n}^{-1}u\right)du$

Apply integration by parts

$u=ta{n}^{-1}u,v^{\prime }=u$

$=2(\frac{1}{2}{u}^2\arctan \left(u\right)-\int \frac{u^2}{2(u^2+1)}du)$

$=2(\frac{1}{2}{u}^2\arctan \left(u\right)-\frac{1}{2}(-\arctan \left(u\right)+u))$

$=2(\frac{1}{2}{\sin }^2\left(x\right)\arctan \left(\sin \left(x\right)\right)-\frac{1}{2}(-\arctan \left(\sin \left(x\right)\right)+\sin \left(x\right)))$

$={\sin }^2\left(x\right)\arctan \left(\sin \left(x\right)\right)+\arctan \left(\sin \left(x\right)\right)-\sin \left(x\right)$

$=\frac{\pi }{2}-1-0$

$=\frac{\pi }{2}-1$