Evaluate the integral: $\int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx$

Let $t = \sin^{-1} x$.
Then,differentiating both sides with respect to $x$,we get $\frac{dt}{dx} = \frac{1}{\sqrt{1-x^2}}$,which implies $dt = \frac{1}{\sqrt{1-x^2}} \, dx$.
Substituting these into the integral,we have:
$\int \frac{\sin^{-1} x}{\sqrt{1-x^2}} \, dx = \int t \, dt$.
Integrating $t$ with respect to $t$ gives $\frac{t^2}{2} + C$.
Substituting back $t = \sin^{-1} x$,the final result is $\frac{(\sin^{-1} x)^2}{2} + C$,where $C$ is an arbitrary constant.