Evaluate the following integrals:

$y=\int \cos 2 x+x^{2} \cos 2 x d x$

$A=\int \cos 2 x d x$

$A=\frac{\sin 2 x}{2}+c_{1}$

$B=\int x^{2} \cos 2 x d x$

Use the method of integration by parts

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

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

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

$B=x^{2} \frac{\sin 2 x}{2}+x \frac{\cos 2 x}{2}-\frac{\sin 2 x}{4}+c_{2}$

The final answer is $\mathrm{y}=\mathrm{A}+\mathrm{B}$

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

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