Evaluate the following integrals:

Let $I=\int x \sin 2 x d x$

Using integration by parts,

$=x \int \sin 2 x d x-\int \frac{d}{d x} x \int \sin 2 x d x$

We know that, $\int \sin n x=\frac{-\cos n x}{n}$ and $\int \cos n x=\frac{\sin n x}{n}$

$=\frac{x}{2}-\cos 2 x+\int \frac{\cos 2 x d x}{2}$

$=-\frac{x}{2} \cos 2 x+\frac{1}{2} \frac{\sin 2 x}{2}+c$

$=-\frac{x}{2} \cos 2 x+\frac{1}{4} \sin 2 x+c$