Answer
$$
\int_{0}^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x =\frac{\pi^{2}}{72}
$$
Work Step by Step
Let $u=\sin ^{-1} x $, so $d u=\frac{d x}{\sqrt{1-x^{2}}} $. When $x=0, u=0 ; $; when $x=\frac{1}{2}, u=\frac{\pi}{6} . $ Thus, the Substitution Rule gives
$$
\begin{aligned}
\int_{0}^{1 / 2} \frac{\sin ^{-1} x}{\sqrt{1-x^{2}}} d x &=\int_{0}^{\pi / 6} u d u \\
&=\left[\frac{u^{2}}{2}\right]_{0}^{\pi / 6} \\
&=\frac{\pi^{2}}{72}
\end{aligned}
$$
