Question

Evaluate ${\int }_{-2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx$.

• A. $0$
• B. $-1$
• C. $1$
• D. $2$

Answer 1

The correct option is A $0$

$I={\int }_{2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx$ ...... $\left(1\right)$

Applying property of limits of integral formula ;

$\Rightarrow I={\int }_{2\pi }^{5\pi }{\cot }^{-1}\left(\tan \right(7\pi -x\left)\right)dx$

$\Rightarrow I={\int }_{2\pi }^{5\pi }{\cot }^{-1}(-\tan x)dx$

$\Rightarrow I=-{\int }_{2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx$ ...... $\left(2\right)$

Adding both equations;

$\Rightarrow 2I=0$

$\Rightarrow I=0$

$\Rightarrow {\int }_{2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx=0$