Question

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

Answer 1

$\begin{array}{c}{\int }_0^1\frac{{\left({\sin }^{-1}x\right)}^3}{\sqrt{1-x^2}}dx\left[\begin{array}{c}x=0=t=0\\ x=1=t=\frac{\pi }{2}\end{array}\right]\\ let,{\sin }^{-1}x=t\\ \frac{1}{\sqrt{1-x^2}}dx=dt\\ {\int }_0^{\frac{\pi }{2}}{t}^{3}dt={\left[\frac{t^4}{4}\right]}_0^{\frac{\pi }{2}}\\ =\frac{{\pi }^4}{16\times 4}=\frac{{\pi }^4}{64}\end{array}$

Hence, this is the answer.