## Calculus: Early Transcendentals (2nd Edition)

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