Answer
\[ = \frac{\pi }{6}\]
Work Step by Step
\[\begin{gathered}
\int_0^{5/2} {\frac{{dx}}{{\sqrt {25 - {x^2}} }}} \hfill \\
\hfill \\
the\,\,integral\,\,has\,\,the\,\,form\,\,{a^2} - {x^2}\,\, \hfill \\
\hfill \\
x = 5\sin \theta \,\,\,\,\,then\,\,\,\,dx = 5\cos \theta d\theta \hfill \\
\hfill \\
and\,\,\sqrt {25 - {x^2}} = 5\cos \theta \hfill \\
\hfill \\
\int_{}^{} {\frac{{5\cos \theta d\theta }}{{5\cos \theta }}} = \int_{}^{} {d\theta } \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \theta + C \hfill \\
\hfill \\
where\,\,\sin \theta = \frac{x}{5} \to \theta = {\sin ^{ - 1}}\,\left( {\frac{x}{5}} \right) \hfill \\
\hfill \\
= \,\,\left[ {{{\sin }^{ - 1}}\,\left( {\frac{x}{5}} \right)} \right]_0^{5/2} \hfill \\
\hfill \\
evaluate\,\,the\,limits \hfill \\
\hfill \\
= {\sin ^{ - 1}}\,\left( {\frac{{5/2}}{5}} \right) - {\sin ^{ - 1}}\,\left( {\frac{0}{5}} \right) \hfill \\
\hfill \\
= \frac{\pi }{6} \hfill \\
\end{gathered} \]
