Question

Integrate the function $x\sin 3x$

Answer 1

Let $I=\int x\sin 3xdx$
Using by parts,
$I=x\int \sin 3xdx-\int \{\left(\frac{d}{dx}x\right)\int \sin 3xdx\}$
$=x\left(\frac{-\cos 3x}{3}\right)-\int 1\cdot \left(\frac{-\cos 3x}{3}\right)dx$
$=\frac{-x\cos 3x}{3}+\frac{1}{3}\int \cos 3xdx$
$=\frac{-x\cos 3x}{3}+\frac{1}{9}\sin 3x+C$