Answer
\[ = \frac{\pi }{2} - 1\]
Work Step by Step
\[\begin{gathered}
\int_0^{\sqrt 2 } {\frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}} dx \hfill \\
\hfill \\
the\,\,integral\,\,has\,\,the\,\,form\,\,{a^2} - {x^2}\,\, \hfill \\
x = 2\sin \theta \,\,\,\,\, \to \,\,\,\,dx = 2\cos \theta d\theta \hfill \\
and\,\,\sqrt {4 - {x^2}} = 2\cos \theta \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_0^{\sqrt 2 } {\frac{{{x^2}}}{{\sqrt {4 - {x^2}} }}} dx = \int_{}^{} {\frac{{\,{{\left( {2\sin \theta } \right)}^2}}}{{2\cos \theta }}\,\left( {2\cos \theta } \right)d\theta } \hfill \\
\hfill \\
= \int_{}^{} {4{{\sin }^2}\theta d\theta } = \hfill \\
\hfill \\
use\,\,{\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2} \hfill \\
\hfill \\
4\int_{}^{} {\,\left( {\frac{{1 - \cos 2\theta }}{2}} \right)} \,d\theta \hfill \\
\hfill \\
= 2\int_{}^{} {\,\left( {1 - \cos 2\theta } \right)d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= 2\theta - \sin \theta + C \hfill \\
\hfill \\
= 2\theta - 2sin\,\theta \cos \,\,\theta + C \hfill \\
\hfill \\
= \,\,\,\left[ {2{{\sin }^{ - 1}}\,\left( {\frac{x}{2}} \right) - 2\,\left( {\frac{x}{2}} \right)\,\left( {\frac{{\sqrt {4 - {x^2}} }}{2}} \right)} \right]_0^{\sqrt 2 } \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \,\,\left[ {2{{\sin }^{ - 1}}\,\left( {\frac{x}{2}} \right) - \frac{{x\sqrt {4 - {x^2}} }}{2}} \right]_0^{\sqrt 2 } \hfill \\
\hfill \\
evaluate\,\,the\,limits\,\,and\,\,simplify \hfill \\
\hfill \\
= \,\,\left[ {2{{\sin }^{ - 1}}\,\left( {\frac{{\sqrt 2 }}{2}} \right) - \frac{{\sqrt 2 \sqrt {4 - \,{{\left( {\sqrt 2 } \right)}^2}} }}{2}} \right] \hfill \\
\hfill \\
= \frac{\pi }{2} - 1 \hfill \\
\hfill \\
\end{gathered} \]
