Answer
\[ = \frac{1}{2}x - \frac{1}{4}\sin 2x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sin }^2}xdx} \hfill \\
\hfill \\
use\,\,the\,identity\,\,{\sin ^2}x = \frac{{1 - \cos 2x}}{2} \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
\int_{}^{} {\,\left( {\frac{{1 - \cos 2x}}{2}} \right)\,\,dx} \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{1}{2}x - \frac{1}{4}\sin 2x + C \hfill \\
\end{gathered} \]