Answer
$$\int_{0}^{\frac{\sqrt{2}}{2}}\frac{x^{2}}{\sqrt{1-x^{2}}}dx=\frac{\pi}{8}-\frac{1}{4}$$
Work Step by Step
$$let\,x=sin\,t,\,dx=cos\,t\,dt(t\in [0,\frac{\pi}{4}])$$
$$\int_{0}^{\frac{\sqrt{2}}{2}}\frac{x^{2}}{\sqrt{1-x^{2}}}dx=\int_{0}^{\frac{\pi}{4}}\frac{sin^{2}t\,cos\,t}{\sqrt{1-sin^{2}t}}dt$$
$$=\int_{0}^{\frac{\pi}{4}}sin^{2}t\,dt=\int_{0}^{\frac{\pi}{4}}\frac{1-cos\,2t}{2}dt$$
$$=\frac{1}{2}\left [t-\frac{1}{2}sin\,2t \right ]_{0}^{\frac{\pi}{4}}$$
$$=\frac{\pi}{8}-\frac{1}{4}$$
