Answer
$$
\int \frac{d x}{\sqrt{1-x^{2}} \sin ^{-1} x}=\ln \left|\sin ^{-1} x\right|+C
$$
where $C$ is an arbitrary constant.
Work Step by Step
$$
\int \frac{d x}{\sqrt{1-x^{2}} \sin ^{-1} x}
$$
Let $u=\sin ^{-1} x$, Then $d u=\frac{1}{\sqrt{1-x^{2}}} d x $. and, the Substitution Rule gives
$$
\begin{aligned}
\int \frac{d x}{\sqrt{1-x^{2}} \sin ^{-1} x}&=\int \frac{1}{u} d u\\
&=\ln |u|+C \\
&=\ln \left|\sin ^{-1} x\right|+C
\end{aligned}
$$
where $C$ is an arbitrary constant.
