## Calculus: Early Transcendentals (2nd Edition)

$= 32{\sin ^{ - 1}}\,\left( {\frac{x}{8}} \right) + \frac{1}{2}x\sqrt {64 - {x^2}} + C$
$\begin{gathered} \int_{}^{} {\sqrt {64 - {x^2}} dx} \hfill \\ \hfill \\ the\,\,integral\,\,has\,\,the\,\,form\,\,{a^2} - {x^2}\,\, \hfill \\ \hfill \\ x = 8\sin \theta \,\,\,\,then\,\,\,\,dx = 8\cos \theta d\theta \hfill \\ and\,\,\sqrt {64 - {x^2}} = 8\cos \theta \hfill \\ \hfill \\ therefore \hfill \\ \hfill \\ \int_{}^{} {\sqrt {64 - {x^2}} dx} = \int_{}^{} {\,\left( {8\cos \theta } \right)\,\left( {8\cos \theta } \right)d\theta } \hfill \\ \hfill \\ simplify \hfill \\ \hfill \\ = 64\int_{}^{} {{{\cos }^2}\theta d\theta } \hfill \\ \hfill \\ use\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \hfill \\ \hfill \\ = 64\int_{}^{} {\,\left( {\frac{{1 + \cos 2\theta }}{2}} \right)d\theta } \hfill \\ \hfill \\ = 32\int_{}^{} {\,\left( {1 + \cos 2\theta } \right)d\theta } \hfill \\ \hfill \\ integrate \hfill \\ \hfill \\ = 32\,\,\left[ {\theta + \frac{{\sin 2\theta }}{2}} \right] + C \hfill \\ \hfill \\ = 32\theta + 16\sin 2\theta + C \hfill \\ \hfill \\ use\,\,\sin 2\theta = 2\sin \theta \cos \theta \hfill \\ \hfill \\ = 32\theta + 16\,\left( {2\sin \theta \cos \theta } \right) + C \hfill \\ \hfill \\ = 32\theta + 32\sin \theta \cos \theta + C \hfill \\ \hfill \\ \hfill \\ substitute\,\,back \hfill \\ = 32{\sin ^{ - 1}}\,\left( {\frac{x}{8}} \right) + 32\,\left( {\frac{x}{8}} \right)\,\left( {\frac{{\sqrt {64 - {x^2}} }}{8}} \right) + C \hfill \\ \hfill \\ = 32{\sin ^{ - 1}}\,\left( {\frac{x}{8}} \right) + \frac{1}{2}x\sqrt {64 - {x^2}} + C \hfill \\ \hfill \\ \end{gathered}$