Answer
\[ = \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {{{\sin }^2}x{{\cos }^2}x} \,dx \hfill \\
\hfill \\
{\text{Using}}\,\,{\text{the }}\,{\text{identities}} \hfill \\
\hfill \\
{\sin ^2}\theta = \frac{{1 - \cos 2\theta }}{2}\,\,and\,\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{1 - \cos \,2x}}{2}} \right)\,\left( {\frac{{1 + \cos 2x}}{2}} \right)dx} \hfill \\
\hfill \\
multiply\,\,and\,distribute \hfill \\
\hfill \\
= \int_{}^{} {\,\left( {\frac{{1 - {{\cos }^2}2x}}{4}} \right)dx} \hfill \\
\hfill \\
= \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {{{\cos }^2}2xdx} \hfill \\
\hfill \\
use\,the\,half\, - \,angle\,\,\,identity\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2}\,\,,\, \hfill \\
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= \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {\frac{{1 + \cos 4x}}{2}} dx \hfill \\
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or \hfill \\
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= \int_{}^{} {\,\frac{1}{4}dx} - \frac{1}{4}\int_{}^{} {\frac{1}{2}} dx - \frac{1}{4}\int_{}^{} {\frac{{\cos 4x}}{2}} dx \hfill \\
\hfill \\
\hfill \\
= \int_{}^{} {\,\frac{1}{4}dx} - \int {\frac{1}{8}} dx - \frac{1}{{32}}\int {\cos 4x} \left( 4 \right)dx \hfill \\
\hfill \\
integrate\,\,and\,simplify \hfill \\
\hfill \\
\frac{1}{4}x - \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C \hfill \\
\hfill \\
= \frac{1}{8}x - \frac{1}{{32}}\sin 4x + C \hfill \\
\end{gathered} \]
