# Evaluate the following integrals:

Question:

Evaluate the following integrals:

$\int \cot ^{6} x d x$

Solution:

Let $I=\int \cot ^{6} x d x$

$\Rightarrow I=\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{4} x d x$

$\Rightarrow I=\int \cot ^{4} x \operatorname{cosec}^{2} x d x-\int \cot ^{4} x d x$

$\Rightarrow I=\int \cot ^{4} x \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x-1\right) \cot ^{2} x d x$

$\Rightarrow I=\int \cot ^{4} x \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cot ^{2} x\right) d x+\int \cot ^{2} x d x$

$\Rightarrow I=\int \cot ^{4} x \operatorname{cosec}^{2} x d x-\int\left(\operatorname{cosec}^{2} x \cot ^{2} x\right) d x+\int\left(\operatorname{cosec}^{2} x-1\right) d x$

Let $\cot x=t$, then

$\Rightarrow-\operatorname{cosec}^{2} x d x=d t$

$\Rightarrow \mathrm{I}=-\int \mathrm{t}^{4} \mathrm{dt}+\int \mathrm{t}^{2} \mathrm{dt}-\int \mathrm{dt}-\int \mathrm{dx}$

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

$\Rightarrow I=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{3} x}{3}-\cot x-x+c$

Therefore, $\int \cot ^{6} x d x=\Rightarrow I=-\frac{\cot ^{5} x}{5}+\frac{\cot ^{3} x}{3}-\cot x-x+c$