Answer
$$\int\frac{dx}{\sqrt{1-x^2}\sin^{-1}x}dx=\ln|\sin^{-1}x|+C$$
Work Step by Step
$$A=\int\frac{dx}{\sqrt{1-x^2}\sin^{-1}x}dx$$
Let $u=\sin^{-1}x$
Then we have $du=\frac{1}{\sqrt{1-x^2}}dx$.
Substitute into $A$: $$A=\int\frac{1}{u}du$$ $$A=\ln|u|+C$$ $$A=\ln|\sin^{-1}x|+C$$
