Answer
\[\frac{{{x^2}}}{4} + \frac{1}{8}\sin 2{x^2} + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {x{{\cos }^2}\,\left( {{x^2}} \right)dx} = \frac{1}{2}\int_{}^{} {{{\cos }^2}\,\left( {{x^2}} \right)\,\left( {2x} \right)dx} \hfill \\
\hfill \\
or \hfill \\
\hfill \\
= \frac{1}{2}\int_{}^{} {{{\cos }^2}udu} \hfill \\
\hfill \\
where\,\,\,{\cos ^2}\theta = \frac{{1 + \cos 2\theta }}{2} \hfill \\
\hfill \\
then \hfill \\
\hfill \\
= \frac{1}{4}\int_{}^{} {\,\left( {1 + \cos 2u} \right)} du \hfill \\
\hfill \\
integrate \hfill \\
\hfill \\
= \frac{u}{4} + \frac{1}{8}\sin 2u + C \hfill \\
\hfill \\
substitute\,\,u = {x^2} \hfill \\
\hfill \\
\frac{{{x^2}}}{4} + \frac{1}{8}\sin 2{x^2} + C \hfill \\
\end{gathered} \]
