Evaluate the following integrals:

Let $\cos x^{2}=t$

Then $d\left(\cos x^{2}\right)=d t$

Since $d\left(x^{n}\right)=n x^{n-1}$ and $d(\cos x)=-\sin x d x$

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

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

hence $\int x \cos ^{3} x^{2} \sin x^{2} d x=\int t^{3} x-\frac{d t}{2}$

$=-\frac{1}{2} \int \mathrm{t}^{3} \mathrm{dt}$

$=-\frac{1}{2} \times \frac{\mathrm{t}^{4}}{4}+\mathrm{c}$

$=-\frac{1}{8} \cos ^{4} \mathrm{x}^{2}+\mathrm{c}$