Question

${\int }_0^1\frac{{\sin }^{-1}x}{2\sqrt{1-x^2}}dx$

Answer 1

We have,

$I={\int }_0^1\frac{{\sin }^{-1}x}{2\sqrt{1-x^2}}dx$

Let $t={\sin }^{-1}x$

$\frac{dt}{dx}=\frac{1}{\sqrt{1-x^2}}$

$dt=\frac{dx}{\sqrt{1-x^2}}$

Therefore,

$I=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}tdt$

$I=\frac{1}{2}{\left[\frac{t^2}{2}\right]}_0^{\frac{\pi }{2}}$

$I=\frac{1}{4}{\left[t^2\right]}_0^{\frac{\pi }{2}}$

$I=\frac{1}{4}\left[\frac{{\pi }^2}{4}\right]$

$I=\frac{{\pi }^2}{16}$

Hence, this is the answer.