Question

The value of $\int {\cot }^{4}xdx$ is

• A. $\frac{{\cot }^{3}x}{3}+\cot x+x+C$
• B. $\frac{1}{3}\cot x+C$
• C. $-\frac{{\cot }^{3}x}{3}+{\cot }^2+x+C$
• D. $-\frac{{\cot }^{3}x}{3}+\cot x+x+C$

Answer 1

The correct option is D $-\frac{{\cot }^{3}x}{3}+\cot x+x+C$

$I=\int {\cot }^{4}xdx=\int {\cot }^{2}x({\text{cosec}}^{2}x-1)dx=\int {\cot }^{2}x{\text{cosec}}^{2}xdx-\int {\cot }^{2}xdx$
$=\int {\cot }^{2}x{\text{cosec}}^{2}xdx-\int ({\text{cosec}}^{2}x-1)dx$
Let $\cot x=t$
$\Rightarrow -{\text{cosec}}^{2}xdx=dt$
$\Rightarrow I=\int t^2(-dt)-\int ({\text{cosec}}^{2}x-1)dx\Rightarrow I=-\frac{t^3}{3}+\cot x+x+C=-\frac{{\cot }^{3}x}{3}+\cot x+x+C$