Question

${\int }_{-1}^{+1}\frac{d}{dx}\left[{\cot }^{-1}\frac{1}{x}\right]dx$ is equals to

• A. $\pi /2$
• B. $-\pi /2$
• C. $\pi$
• D. $-\pi$

Answer 1

The correct option is B $-\pi /2$

Now,

$\underset{-1}{\overset{1}{\int }}\frac{d}{dx}\left[{\cot }^{-1}\frac{1}{x}\right]dx$

$=\underset{-1}{\overset{1}{\int }}d\left[{\cot }^{-1}\frac{1}{x}\right]$

$={\left[{\cot }^{-1}\frac{1}{x}\right]}_{-1}^1$

$={\cot }^{-1}\left(1\right)-{\cot }^{-1}(-1)$

$=\frac{\pi }{4}-\frac{3\pi }{4}$

$=-\frac{\pi }{2}$.