Evaluate the following integrals:

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

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

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

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

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

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

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

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

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

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

Put back $\mathrm{t}=\sin \mathrm{x}$

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