Question

The value of ${\int }_{-2\pi }^{5\pi }{\cot }^{-1}\left(\tan x\right)dx$ is equal to:

• A. $\frac{7\pi }{2}$
• B. $\frac{7{\pi }^2}{2}$
• C. $\frac{3\pi }{2}$
• D. None of these

Answer 1

The correct option is D None of these

Let ${\cot }^{-1}\left(\tan x\right)=\theta$

Then, $\tan x=\cot \theta$

Differentiating,

${\sec }^{2}xdx=-{\csc }^2\theta d\theta$

$dx=-\frac{{\csc }^2\theta }{{\sec }^{2}x}d\theta$

Since

$\tan x=\cot \theta$

$\Rightarrow \pm \sqrt{{\sec }^{2}x-1}=\cot \theta$

${\sec }^{2}x-1={\cot }^2\theta$

${\sec }^{2}x=1+{\cot }^2\theta ={\csc }^2\theta$

Hence,

$dx=-\frac{{\csc }^2\theta }{{\sec }^{2}x}d\theta =-d\theta$

$\therefore \int {\cot }^{-1}\left(\tan x\right)dx=-\int \theta d\theta$

$=-\frac{{\theta }^2}{2}+c$

Since the function $y={\cot }^{-1}x$ has the range $(0,\pi )$,

${\cot }^{-1}\left(\tan \right(-2\pi \left)\right)={\cot }^{-1}\left(0\right)=\frac{\pi }{2}$

${\cot }^{-1}\left(\tan \right(5\pi \left)\right)={\cot }^{-1}\left(0\right)=\frac{\pi }{2}$

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