Question

Integrate the function $\frac{{\sin }^{-1}x}{\sqrt{1-x^2}}$

Answer 1

Let ${\sin }^{-1}x=t$
$\therefore \frac{1}{\sqrt{1-x^2}}dx=dt$
$\Rightarrow \int \frac{{\sin }^{-1}x}{\sqrt{1-x^2}}dx=\int tdt$
$=\frac{t^2}{2}+C$
$=\frac{({\sin }^{-1}x{)}^2}{2}+C$