Answer
$$\int x\sin^2xdx=\frac{x^2}{4}-\frac{1}{4}x\sin2x-\frac{1}{8}\cos2x+C$$
Work Step by Step
$$A=\int x\sin^2xdx=\int x\Big(\frac{1-\cos2x}{2}\Big)dx$$ $$A=\frac{1}{2}\int xdx-\frac{1}{2}\int x\cos2xdx$$ $$A=\frac{x^2}{4}-\frac{1}{2}\int x\cos2xdx$$
Set $u= x$ and $dv=\cos2xdx$
We would have $du=dx$ and $v=\frac{1}{2}\sin2x$
Applying Integration by Parts, we have
$$A=\frac{x^2}{4}-\frac{1}{2}\Big(\frac{1}{2}x\sin2x-\frac{1}{2}\int\sin2xdx\Big)$$ $$A=\frac{x^2}{4}-\frac{1}{2}\Big(\frac{1}{2}x\sin2x+\frac{1}{4}\cos2x\Big)+C$$ $$A=\frac{x^2}{4}-\frac{1}{4}x\sin2x-\frac{1}{8}\cos2x+C$$
