Evaluate the following integrals:

Solution:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Put back $t=\cos x$

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

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

$=-\cos x+\frac{2}{3} \cos ^{3} x-\frac{1}{5} \cos ^{5} x+c=-\left[\cos x-\frac{2}{3} \cos ^{3} x+\frac{1}{5} \cos ^{5} x\right]+c$