Question

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

Answer 1

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

Let $t={\sin }^{-1}x\Rightarrow dt=\frac{1}{\sqrt{1-x^2}}dx$

$={\int }_0^{1}tdt$

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

$=\frac{1}{2}{\left[{\left({\sin }^{-1}x\right)}^2\right]}_0^1$

$=\frac{1}{2}[{\left({\sin }^{-1}1\right)}^2-{\left({\sin }^{-1}0\right)}^2]$

$=\frac{1}{2}[{\left(\frac{\pi }{2}\right)}^2-0]$

$=\frac{{\pi }^2}{8}$