Answer
\[ = {\sin ^{ - 1}}\,\left( {\frac{x}{6}} \right) + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {\frac{{dx}}{{\sqrt {36 - {x^2}} }}} \hfill \\
\hfill \\
the\,\,integral\,\,contains\,\,{a^2} - {x^2}\,\,so \hfill \\
\hfill \\
x = 6\sin \theta \,\,\,\,then\,\,\,dx = 6\cos \theta d\theta \hfill \\
and\,\,\,\sqrt {36 - {x^2}} = 6\cos \theta \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\frac{{dx}}{{\sqrt {36 - {x^2}} }}} = \int_{}^{} {\frac{{6\cos \theta d\theta }}{{6\cos \theta }}} \hfill \\
\hfill \\
simplify \hfill \\
\hfill \\
= \int_{}^{} {d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \theta + C \hfill \\
\hfill \\
where\,\,\,\,\,\theta = {\sin ^{ - 1}}\,\left( {\frac{x}{6}} \right) \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= {\sin ^{ - 1}}\,\left( {\frac{x}{6}} \right) + C \hfill \\
\hfill \\
\end{gathered} \]
