Evaluate the following integrals:

$y=\int\left(1-\cos ^{2} x\right) \cos ^{4} x \sin x d x$'

Let, $\cos x=t$

Differentiating both side with respect to $x$

$\frac{d t}{d x}=-\sin x \Rightarrow-d t=\sin x d x$

$y=\int-\left(1-t^{2}\right) t^{4} d t$

$y=-\int t^{4}-t^{6} d t$

Using formula $\int t^{n} d t=\frac{t^{n+1}}{n+1}$

$y=-\left(\frac{t^{5}}{5}-\frac{t^{7}}{7}\right)+c$

Again, put $t=\cos x$

$y=\frac{\cos ^{7} x}{7}-\frac{\cos ^{5} x}{5}+c$