Answer
$$ - \frac{1}{2}\pi $$
Work Step by Step
$$\eqalign{
& \int_0^\pi {x\sin 2x} dx \cr
& {\text{Integrate by parts}} \cr
& {\text{Let }}u = x,{\text{ }}du = dx \cr
& dv = \sin 2xdx,{\text{ }}v = - \frac{1}{2}\cos 2x \cr
& {\text{Integration by parts formula }} \cr
& \int {\underbrace x_u} \underbrace {\sin 2xdx}_{dv} = \underbrace {\left( x \right)}_u\underbrace {\left( { - \frac{1}{2}\cos 2x} \right)}_v - \int {\underbrace {\left( { - \frac{1}{2}\cos 2x} \right)}_v} \underbrace {dx}_{du} \cr
& {\text{Multiply}} \cr
& \int {x\sin 2x} dx = - \frac{1}{2}x\cos 2x + \frac{1}{2}\int {\cos 2x} dx \cr
& \int {x\sin 2x} dx = - \frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x + C \cr
& {\text{Therefore,}} \cr
& \int_0^\pi {x\sin 2x} dx = \left[ { - \frac{1}{2}x\cos 2x + \frac{1}{4}\sin 2x} \right]_0^\pi \cr
& = \left[ { - \frac{1}{2}\left( \pi \right)\cos 2\left( \pi \right) + \frac{1}{4}\sin 2\left( \pi \right)} \right] - \left[ { - 0 + \frac{1}{4}\sin 0} \right] \cr
& {\text{Simplify}} \cr
& = \left( { - \frac{1}{2}\pi + 0} \right) - \left( 0 \right) \cr
& = - \frac{1}{2}\pi \cr} $$
