Answer
$$\frac{\pi -2}{8}$$
Work Step by Step
Given $$\int_{0}^{\sqrt{2}/2}\frac{x^2}{\sqrt{1-x^2}}dx$$
Let $x=\sin u\ \ \Rightarrow \ \ dx =\cos udu$ , at $x=0\to u=0$ , at $x=\sqrt{2}/2\to u=\pi/4$ , then
\begin{align*}
\int_{0}^{\sqrt{2}/2}\frac{x^2}{\sqrt{1-x^2}}dx&=\int_{0}^{\pi/4}\frac{\sin^2u\cos udu}{\sqrt{1-\sin^2u}} \\
&=\int_{0}^{\pi/4} \sin^2u du \\
&=\frac{1}{2}\int_{0}^{\pi/4} (1-\cos2u) du\\
&=\frac{1}{2}\left[ u- \frac{1}{2}\sin 2u\right]_{0}^{\pi/4}\\
&=\frac{1}{2}\left[\frac{\pi}{4}- \frac{1}{2}\right]\\
&=\frac{\pi -2}{8}
\end{align*}