Question

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

Answer 1

We have,

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

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

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

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

Therefore,

$I=\int tdt$

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

On putting the value of $t$, we get

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

Hence, this is the answer.