Evaluate the following integrals:

$\sin ^{5} x=\sin ^{4} x \cdot \sin x$

Assume $\cos x=t$

$d(\cos x)=d t$

$-\sin x \cdot d x=d t$

$\Rightarrow \mathrm{dx}=\frac{-\mathrm{dt}}{\sin \mathrm{x}}$

Substitute $t$ and $d t$ we get

$\Rightarrow \int \frac{\sin ^{4} x \cdot \sin x}{\cos ^{4} x} \times \frac{-d t}{\sin x}$

$\Rightarrow \int \frac{-d t\left(1-\cos ^{2} x\right)^{2}}{\cos ^{4} x}$

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

$\Rightarrow-\int \frac{1+t^{4}-2 t^{2}}{t^{4}} d t$

$\Rightarrow-\int \frac{1}{t^{4}} d t-\int \frac{t^{4}}{t^{4}} d t+2 \int \frac{t^{2}}{t^{4}} d t$

$\Rightarrow-\int t^{-4} d t-\int d t+2 \int t^{-2} d t$

But $t=\cos x$

$\Rightarrow \frac{\cos ^{-2} x}{3}-\cos x-2 \cos ^{-1} x+c$