## Calculus: Early Transcendentals (2nd Edition)

$= \frac{9}{4}{\sin ^{ - 1}}\,\left( {\frac{{2x}}{3}} \right) + \frac{1}{2}x\sqrt {9 - 4{x^2}} + C$
$\begin{gathered} \int_{}^{} {\sqrt {9 - 4{x^2}} dx} \hfill \\ \hfill \\ rewrite \hfill \\ \hfill \\ = \int_{}^{} {\sqrt {\,{{\left( 3 \right)}^2} - \,{{\left( {2x} \right)}^2}} dx} \hfill \\ \hfill \\ the\,\,\,integral\,has\,\,the\,\,form\,\,{x^2} - {a^2}\,\, \hfill \\ \hfill \\ 2x = 3\sin \theta \,\,\,\,\,\,\, \to \,\,\,x = \frac{3}{2}\sin \theta \,\, \to \,dx = \frac{3}{2}\cos \theta d\theta \,\, \hfill \\ and\,\,\,\sqrt {\,{{\left( 3 \right)}^2} - \,\left( {2{x^2}} \right)} = 3\cos \theta \hfill \\ \hfill \\ apply\,\,the\,\,substitution \hfill \\ \hfill \\ = \int_{}^{} {\,\left( {3\cos \theta } \right)\,\left( {\frac{3}{2}} \right)\cos \theta d\theta \,} = \frac{9}{2}\int_{}^{} {{{\cos }^2}\theta d\theta } \hfill \\ \hfill \\ use\,trigonometric\,\,identity{\text{ }}{\cos ^2}\theta = \frac{{1 + \cos \theta }}{2} \hfill \\ \hfill \\ = \frac{9}{2}\int_{}^{} {\,\left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = \frac{9}{4}\theta + \frac{9}{8}\sin 2\theta + C \hfill \\ \hfill \\ = \frac{9}{4}\theta + \frac{9}{4}\cos \theta \sin \theta + C \hfill \\ \hfill \\ substitute\,\,for\,\,\sin \theta {\text{ and cos}}\theta \hfill \\ \hfill \\ = \frac{9}{4}{\sin ^{ - 1}}\,\left( {\frac{{2x}}{3}} \right) + \frac{1}{2}x\sqrt {9 - 4{x^2}} + C \hfill \\ \end{gathered}$