Question

${\int }_0^{\pi /2}\frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx$

Answer 1

We have,

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

Let

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

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

Therefore,

$I={\int }_0^1{t}^{3}dt$

$\Rightarrow {{\left[\frac{t^4}{4}\right]}_0}^1$

$\Rightarrow \left[\frac{1-0}{4}\right]$

$\Rightarrow \frac{1}{4}$

Hence, this is the answer.