Find the reduction formula for $\int {\cot }^{\mathrm{n}}\mathrm{dx}$ and n being positive integer $\mathrm{n}\ge 2$ and hence deduce the value of $\int {\cot }^4\mathrm{xdx}$
$\text{ If }{\mathrm{I}}_{\mathrm{n}}=\int {\cot }^{\mathrm{n}}\mathrm{xdx} then {\mathrm{I}}_{\mathrm{n}}=\frac{-{\cot }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1}-{\mathrm{I}}_{\mathrm{n}-2}$

$\begin{array}{l}{\mathrm{I}}_{\mathrm{n}}=\int {\cot }^{\mathrm{n}-2}{\mathrm{xcot}}^2\mathrm{xdx}\\ =\int {\cot }^{\mathrm{n}-2}\mathrm{x}\left({\mathrm{cosec}}^2\mathrm{x}-1\right)\mathrm{dx}\\ =\int {\cot }^{\mathrm{n}-2}{\mathrm{xcosec}}^2\mathrm{xdx}-\int {\cot }^{\mathrm{n}-2}\mathrm{xdx}\\ =-\frac{{\cot }^{\mathrm{n}-1}\mathrm{x}}{\mathrm{n}-1}-{\mathrm{I}}_{\mathrm{n}-2}\\ {\mathrm{I}}_4=\int {\cot }^4\mathrm{xdx}=\frac{-{\cot }^3\mathrm{x}}{3}-{\mathrm{I}}_2\\ =\frac{-{\cot }^3\mathrm{x}}{3}-\left[\frac{-\cot \mathrm{x}}{1}-{\mathrm{I}}_0\right]\\ \therefore {\mathrm{I}}_4=\frac{-{\cot }^3\mathrm{x}}{3}+\cot \mathrm{x}+\mathrm{x}+\mathrm{c}\end{array}$