Question

Integrate : $\int x{\sin }^{2}x$

Answer 1

$\int x{\sin }^{2}xdx=\int x\left[\frac{1}{2}\right(1-\cos 2x\left)\right]dx(\because \cos 2x=1-2{\sin }^{2}x)=\frac{1}{2}\int xdx-\frac{1}{2}\int x\cos 2xdx=\frac{x^2}{4}-\frac{1}{2}[x\int \cos 2xdx-\int (\frac{dx}{dx}.\int \cos 2xdx)]dx=\frac{x^2}{4}-\frac{x\sin 2x}{4}-\frac{\cos 2x}{8}+C=\frac{{2x}^2-2x\sin 2x-\cos 2x}{8}+C$