Question

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

Answer 1

$I=\int {\cot }^{4}x$

$=\int {\cot }^{2}x({\text{cosec}}^{2}x-1)dx$

$=\int {\cot }^{2}x\times {\text{cosec}}^{2}xdx-\int {\cot }^{2}xdx$

$=\int {\cot }^{2}x\times {\text{cosec}}^{2}xdx-\int ({\text{cosec}}^{2}x-1)dx$

$=I_1-I_2$

$I_1=\int {\cot }^{2}x\times {\text{cosec}}^{2}xdx$

put,

$\cot x=t$

$-{\text{cosec}}^{2}xdx=dt$

$=-\int t^{2}dt$

$=-\frac{t^3}{3}$

$=-\frac{{\cot }^{3}x}{3}+C$

then,

$I_2$=$\int ({\text{cosec}}^{2}x-1)dx$

$=-(\cot x+x)+C$

now,

$I=I_1-I_2$

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