Answer
$$\int \frac{\sqrt{\sin^{-1}x}dx}{\sqrt{1-x^2}}=\frac{2(\sin^{-1}x)^{3/2}}{3}+C$$
Work Step by Step
$$A=\int \frac{\sqrt{\sin^{-1}x}dx}{\sqrt{1-x^2}}$$
We set $a=\sin^{-1}x$, which means $$da=\frac{dx}{\sqrt{1-x^2}}$$
Therefore, $$A=\int \sqrt ada=\int a^{1/2}da$$ $$A=\frac{a^{3/2}}{\frac{3}{2}}+C$$ $$A=\frac{2a^{3/2}}{3}+C$$ $$A=\frac{2(\sin^{-1}x)^{3/2}}{3}+C$$
