## Calculus, 10th Edition (Anton)

$$- \frac{\pi }{2}$$
\eqalign{ & \int_0^\pi {x\sin 2x} dx \cr & {\text{substitute }}u = x,{\text{ }}du = dx \cr & dv = \sin 2xdx,{\text{ }}v = - \frac{1}{2}\cos 2x \cr & {\text{applying integration by parts}}{\text{, we have}} \cr & \int_0^\pi {x\sin 2x} dx = \left. {\left( { - \frac{x}{2}\cos 2x} \right)} \right|_0^\pi - \int_0^\pi {\left( { - \frac{1}{2}\cos 2x} \right)dx} \cr & {\text{integrating}} \cr & \int_0^\pi {x\sin 2x} dx = \left. {\left( { - \frac{x}{2}\cos 2x} \right)} \right|_0^\pi + \left. {\left( {\frac{1}{4}\sin 2x} \right)} \right|_0^\pi \cr & \int_0^\pi {x\sin 2x} dx = \left. {\left( {\frac{x}{2}\cos 2x + \frac{1}{4}\sin 2x} \right)} \right|_0^\pi \cr & {\text{evaluate limits}} \cr & = \left( { - \frac{\pi }{2}\cos 2\pi + \frac{1}{4}\sin 2\pi } \right) - \left( { - \frac{0}{2}\cos 0 + \frac{1}{4}\sin 0} \right) \cr & {\text{simplify}} \cr & = - \frac{\pi }{2}\left( 1 \right) + \frac{1}{4}\left( 0 \right) \cr & = - \frac{\pi }{2} \cr}