Question:
Evaluate the following integrals:
$\int \cot ^{5} x \operatorname{cosec}^{4} x d x$
Solution:
Let $I=\int \cot ^{5} x \operatorname{cosec}^{4} x d x$
$\Rightarrow I=\int \cot ^{5} x \operatorname{cosec}^{2} x \operatorname{cosec}^{2} x d x$
$\Rightarrow I=\int \cot ^{5} x\left(1+\cot ^{2} x\right) \operatorname{cosec}^{2} x d x$
$\Rightarrow I=\int\left(\cot ^{5} x+\cot ^{7} x\right) \operatorname{cosec}^{2} x d x$
Let $\cot x=t$, then
$\Rightarrow-\operatorname{cosec}^{2} x d x=d t$
$\Rightarrow I=-\int\left(t^{5}+t^{7}\right) d t$
$\Rightarrow I=-\frac{t^{6}}{6}-\frac{t^{8}}{8}+c$
$\Rightarrow I=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+c$
Therefore, $\int \cot ^{5} x \operatorname{cosec}^{4} x d x=-\frac{\cot ^{6} x}{6}-\frac{\cot ^{8} x}{8}+c$