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