Evaluate the following integrals:

Solution:

$\int \cos ^{5} x d x=\int \cos ^{3} x \cos ^{2} x d x$

$=\int \cos ^{3} x\left(1-\sin ^{2} x\right) d x\left\{\right.$ since $\left.\sin ^{2} x+\cos ^{2} x=1\right\}$

$=\int\left(\cos ^{3} x-\cos ^{3} x \sin ^{2} x\right) d x$

$=\int\left(\cos x\left(\cos ^{2} x\right)-\cos ^{3} x \sin ^{2} x\right) d x$

$=\int\left(\cos x\left(1-\sin ^{2} x\right)-\cos ^{3} x \sin ^{2} x\right) d x\left\{\right.$ since $\left.\sin ^{2} x+\cos ^{2} x=1\right\}$

$=\int\left(\cos x-\cos x \sin ^{2} x-\cos ^{3} x \sin ^{2} x\right) d x$

$=\int \cos x d x-\int \cos x \sin ^{2} x d x-\int \cos ^{3} x \sin ^{2} x d x$ (separate the integrals)

We know, $d(\sin x)=\cos x d x$

So put $\sin x=t$ and $d t=\cos x d x$ in above integrals

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

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

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

$\left.=\int \cos x d x-\int t^{2} d t\right)-\int\left(1-t^{2}\right) t^{2} d t$

$\left.=\int \cos x d x-\int t^{2} d t\right)-\int\left(t^{2}-t^{4}\right) d t$

$=\sin x-\frac{t^{3}}{3}-\frac{t^{3}}{3}+\frac{t^{5}}{5}+c\left(\right.$ since $\int x^{n} d x=\frac{x^{2+1}}{n+1}+c$ for any $\left.c \neq-1\right)$

Put back $t=\sin x$

$=\sin x-\frac{\sin ^{2} x}{3}-\frac{\sin ^{2} x}{3}+\frac{\cos ^{5} x}{5}+c$

$=\sin x-\frac{2}{3} \sin ^{3} x+\frac{1}{5} \sin ^{5} x+c$