Answer
\[ = - \frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x\, + C\]
Work Step by Step
\[\begin{gathered}
\int_{}^{} {x\sin \,2x\,dx} \hfill \\
\hfill \\
set\,\,\,the\,\,substitution \hfill \\
\hfill \\
u = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,then\,\,\,\,\,\,\,du = dx \hfill \\
du = \sin 2xdx\,\,\,\,\,\,\,then\,\,\,v = - \frac{1}{2}\cos \,2x \hfill \\
\hfill \\
use\,\,uv - \int_{}^{} {vdu} \hfill \\
\hfill \\
{\text{replacing the values }}{\text{in the equation}} \hfill \\
\hfill \\
= - \frac{x}{2}\cos 2x - \int_{}^{} {\,\left( { - \frac{1}{2}\cos 2x} \right)} dx \hfill \\
\hfill \\
= - \frac{x}{2}\cos 2x + \frac{1}{4}\int_{}^{} {\,\left( {\cos 2x} \right)\left( 2 \right)} dx \hfill \\
\hfill \\
\operatorname{int} egrating \hfill \\
\hfill \\
= - \frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x\, + C \hfill \\
\end{gathered} \]
