Question

$\int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx$

Answer 1

$I$ =$\int \frac{si{n}^{-1}x}{\sqrt{1-x^2}}dx$

Let $si{n}^{-1}x=t$

Then,

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

Therefore,

$I$ = $\int tdt$ = $\frac{t^2}{2}+C$ = $\frac{(si{n}^{-1}x{)}^2}{2}+C$