Question

Evaluate: $I={\int }_0^1\frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx$

Answer 1

Let $I={\int }_0^1\frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx$
Put $t={\sin }^{-1}x$
$dt=\frac{1}{\sqrt{1-x^2}}dx$
$\therefore I={\int }_0^{\pi /2}{t}^{3}dt={\left[\frac{t^4}{4}\right]}_0^{\pi /2}$

$=\frac{1}{4}[{\left(\frac{\pi }{2}\right)}^4-0]$
$=\frac{1}{4}\left[\frac{{\pi }^4}{16}\right]=\frac{{\pi }^4}{64}$