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