## Calculus: Early Transcendentals (2nd Edition)

$= 8{\sin ^{ - 1}}\,\left( {\frac{x}{4}} \right) - \frac{1}{2}x\sqrt {16 - {x^2}} + C$
$\begin{gathered} \int_{}^{} {\frac{{{x^2}}}{{\sqrt {16 - {x^2}} }}dx} \hfill \\ \hfill \\ the\,\,\,integral\,contains\,\,the\,\,form\,\,{x^2} - {a^2}\,\, \hfill \\ then \hfill \\ \hfill \\ x = 4\sin \theta \,\,\,\, \to \,\,dx = 4\cos \theta d\theta \hfill \\ and\,\sqrt {16 - {x^2}} = 4\cos \theta \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ \int_{}^{} {\frac{{{x^2}}}{{\sqrt {16 - {x^2}} }}dx} = \int_{}^{} {\frac{{16{{\sin }^2}\theta }}{{4\cos \theta }}} \,\left( {4\cos \theta } \right)d\theta \hfill \\ \hfill \\ use\,trigonometric\,\,identity{\text{ }}{\sin ^2}\theta = \frac{{1 - \cos \theta }}{2} \hfill \\ \hfill \\ = 16\int_{}^{} {{{\sin }^2}\theta d\theta } = 16\int_{}^{} {\,\left( {\frac{{1 - \cos \theta }}{2}} \right)d\theta } \hfill \\ \hfill \\ = 8\int_{}^{} {\,\left( {1 - \cos 2\theta } \right)} \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = 8\,\,\left[ {\theta - \frac{1}{2}\sin 2\theta } \right] + C \hfill \\ \hfill \\ = 8\theta - 4\sin 2\theta + C \hfill \\ \hfill \\ = 8\theta - 8\sin \theta \cos \theta + C \hfill \\ \hfill \\ substitute\,\,for\,\,\sin \theta {\text{ and cos}}\theta \hfill \\ \hfill \\ = 8{\sin ^{ - 1}}\,\left( {\frac{x}{4}} \right) - 8\,\left( {\frac{x}{4}} \right)\,\left( {\frac{{\sqrt {16 - {x^2}} }}{4}} \right) + C \hfill \\ \hfill \\ = 8{\sin ^{ - 1}}\,\left( {\frac{x}{4}} \right) - \frac{1}{2}x\sqrt {16 - {x^2}} + C \hfill \\ \hfill \\ \end{gathered}$