Answer
\[\frac{\pi }{6}\]
Work Step by Step
\[\begin{gathered}
\int_0^{1/2} {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} \hfill \\
\hfill \\
{\text{integrating}}{\text{, }}\frac{d}{{dx}}\left[ {\arcsin x} \right] = \frac{1}{{\sqrt {1 - {x^2}} }} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
\int_0^{1/2} {\frac{{dx}}{{\sqrt {1 - {x^2}} }}} = \left[ {{{\sin }^{ - 1}}\left( x \right)} \right]_0^{1/2} \hfill \\
\hfill \\
{\text{Use the fundamental theorem of calculus}} \hfill \\
\hfill \\
{\sin ^{ - 1}}\left( {\frac{1}{2}} \right) - {\sin ^{ - 1}}\left( 0 \right) \hfill \\
\hfill \\
\frac{\pi }{6} \hfill \\
\end{gathered} \]
