Question

Solve:

$\int \frac{{\left({\sin }^{-1}x\right)}^2}{\sqrt{1-x^2}}dx,x\in (-1,1)$

Answer 1

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

Putting ${\sin }^{-1}x=m$ $\Rightarrow \frac{1}{\sqrt{1-x^2}}dx=dm$

$\Rightarrow I=\int m^{2}dm$

$\Rightarrow I=\frac{m^3}{3}{|}_{-1}^1$

$\Rightarrow I=\frac{{\left({\sin }^{-1}x\right)}^3}{3}{|}_{-1}^1=-{\pi }^3$