Answer
$ -\displaystyle \frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x+C$
Work Step by Step
Use the double angle identity $\sin 2x=2\sin x\cos x$
$I=\displaystyle \frac{1}{2}\int x\sin 2xdx\quad $
now, by parts, $\displaystyle \left[\begin{array}{ll}
u=x & dv=\sin 2x\\
du=dx & dv=-\frac{1}{2}\cos 2xdx
\end{array}\right],\quad\int udv=uv-\int vdu$
$I=\displaystyle \frac{1}{2}(-\frac{1}{2}x\cos 2x-\int-\frac{1}{2}\cos 2xdx)$
$=-\displaystyle \frac{1}{4}x\cos 2x+\frac{1}{4}\cdot\frac{1}{2}\sin 2x+C$
$=-\displaystyle \frac{1}{4}x\cos 2x+\frac{1}{8}\sin 2x+C$
