Question

${\int }_0^1{\sin }^{-1}\left(\frac{2x}{1+x^2}\right)dx$.

Answer 1

Let $x=tan\theta$

$dx=se{c}^2\theta d\theta$

when $x=0,\theta =0$

$x=1,\theta =\frac{\pi }{4}$

${\int }_0^{\frac{\pi }{4}}si{n}^{-1}\left(\frac{2tan\theta }{1+ta{n}^2\theta }\right)se{c}^2\theta d\theta$

${\int }_0^{\frac{\pi }{4}}si{n}^{-1}\left(sin2\theta \right)se{c}^2\theta d\theta$

$2{\int }_0^{\frac{\pi }{4}}\theta se{c}^2\theta d\theta$

Using byparts

$u=\theta ,v=se{c}^2\theta$

$du=1,\int vdv=tan\theta$,we get

$[\theta tan\theta -\int tan\theta d\theta {]}_0^{\frac{\pi }{4}}$

$[\theta tan\theta -log(|cos\theta |){]}_0^{\frac{\pi }{4}}$

$\frac{\pi }{4}+log\sqrt{2}$