Question

Integrate:

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

Answer 1

Consider the given integral.

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

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

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

Therefore,

$I=\int tdt$

$I=\frac{t^2}{2}+C$

On putting the value of $t$, we get

$I=\frac{{\left({\sin }^{-1}x\right)}^2}{2}+C$

Hence, this is the answer.