Question

$\int x{\sin }^{2}xdx=\frac{x^2}{4}-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x.$

Answer 1

As ${\sin }^{2}x=\frac{1-\cos 2x}{2},{\cos }^{2}x=\frac{1+\cos 2x}{2}$
$\therefore I=\int x{\sin }^{2}xdx=\int \frac{x(1-\cos 2x)}{2}dx$
$=\int \frac{x}{2}dx-\frac{1}{2}\int x\cos 2xdx$
Applying integration by parts in second integral
$=\frac{x^2}{4}-\frac{1}{2}[\frac{x\sin 2x}{2}-\int 1.\frac{\sin 2x}{2}dx]$
$=\frac{x^2}{4}-\frac{1}{4}x\sin 2x+\frac{1}{2}[-\frac{\cos 2x}{4}]$
$=\frac{x^2}{4}-\frac{1}{4}x\sin 2x-\frac{1}{8}\cos 2x+C$