Evaluate the following integrals:

$\int \sin ^{7} x d x=\int \sin ^{6} x \sin x d x$

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

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

We know $(a-b)^{3}=a^{3}-b^{3}-3 a^{2} b+3 a b^{2}$

Here, $a=1$ and $b=\cos ^{2} x$

Hence, $\int\left(1-\cos ^{2} x\right)^{3} \sin x d x=\int\left(1-\cos ^{6} x-3 \cos ^{2} x+3 \cos ^{4} x\right) \sin x d x$

$=\int\left(\sin x d x-\cos ^{6} x \sin x d x-3 \cos ^{2} x \sin x d x+3 \cos ^{4} x \sin x d x\right)$ \{take sin $x d x$ inside brackets)

$=\int \sin x d x-\int \cos ^{6} x \sin x d x-3 \int \cos ^{2} x \sin x d x+3 \int \cos ^{4} x \sin x d x$ (separate the integrals)

Put $\cos x=t$ and $-\sin x d x=d t$

$=\int \sin x d x-\int t^{6}(-d t)-3 \int t^{2}(-d t)+3 \int t^{4}(-d t)$

$=-\cos x+\frac{t^{7}}{7}+\frac{3 t^{3}}{3}-\frac{3 t^{5}}{5}+c$

$=-\cos x+\frac{t^{7}}{7}+t^{3}-\frac{3 t^{5}}{5}+c$

Put back $t=\cos x$

$=-\cos x+\cos ^{3} x-\frac{3}{5} \cos ^{5} x+\frac{1}{7} \cos ^{7} x+c$