Evaluate the following integrals:

Let $I=\int x^{2} \cos 2 x d x$

Using integration by parts,

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

We know that,

$\int \cos 2 x d x=\sin 2 x$ and $\frac{d}{d x} x^{2}=2 x$

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

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

Using integration by parts in $\int x \sin 2 x d x$

$=\frac{x^{2}}{2} \sin 2 x-\left(x \int \sin 2 x d x-\int \frac{d}{d x} x \int \sin 2 x d x\right)$

$=\frac{x^{2}}{2} \sin 2 x-\left(\frac{-x}{2} \cos 2 x+\frac{1}{2} \int \cos 2 x d x\right)$

$=\frac{x^{2}}{2} \sin 2 x-\left(\frac{-x}{2} \cos 2 x+\frac{1}{4} \sin 2 x\right)+c$

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