Answer
$$\frac{1}{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right).$$
Work Step by Step
Since $x=\sin \theta $, then $dx=\cos \theta d\theta$, and when $x
:0\to 1/2$ then $\theta
:0\to \pi/6$; hence, we have
$$
\int_{0}^{1 / 2} \frac{x^{2}}{\sqrt{1-x^{2}}} d x=\int_{0}^{\pi / 6} \frac{\sin^2 \theta }{\sqrt{1-\sin^{2}\theta}} d \theta\\
=\int_{0}^{\pi / 6} \frac{\sin^2 \theta }{\sqrt{1-\sin^{2}\theta}} \cos \theta d \theta\\
=\int_{0}^{\pi / 6}\sin^2\theta d\theta=\frac{1}{2}\int_{0}^{\pi / 6}(1-\cos 2\theta ) d\theta\\
=\frac{1}{2}(\theta -\frac{1}{2} \sin2\theta)|_{0}^{\pi / 6}=
=\frac{1}{2}\left(\frac{\pi}{6}-\frac{\sqrt{3}}{4}\right).
$$
