Question

$\int {\cot }^{4}xdx$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int {\cot }^{4}xdx \\ =\int {\cot }^{2}x·{\cot }^{2}xdx \\ =\int {\cot }^{2}x·\left({\mathrm{cosec}}^{2}x-1\right)dx \\ =\int {\cot }^{2}x·{\mathrm{cosec}}^{2}xdx-\int {\cot }^{2}xdx \\ =\int {\cot }^{2}x·{\mathrm{cosec}}^{2}xdx-\int \left({\mathrm{cosec}}^{2}x-1\right)dx \\ \mathrm{Putting}\cot x=t\mathrm{in}\mathrm{the}\mathrm{Ist}\mathrm{integral} \\ \Rightarrow -{\mathrm{cosec}}^{2}xdx=dt \\ \therefore I=-\int t^{2}dt-\int \left({\mathrm{cosec}}^{\mathit{2}}x-1\right)\mathit{}dx \\ =\frac{-t^3}{3}+\cot x+x+C \\ =\frac{-{\cot }^{3}x}{3}+\cot x+x+C\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\mathit{}\left[\because t\mathit{=}\mathit{}\cot \mathit{}x\right] \end{gathered}$$