Question

Evaluate ${\int }_{-1}^1\frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)dx$

• A. $\frac{\pi }{2}$
• B. $\frac{-\pi }{2}$
• C. $\pi$
• D. $-\pi$

Answer 1

The correct option is B $\frac{-\pi }{2}$

${\int }_{-1}^1\frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)dx$

We know that $\Rightarrow \frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)=\frac{1}{1+(\frac{1}{x}{)}^2}\times \left(\frac{-1}{x^2}\right)$

$\Rightarrow \frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)=-\frac{1}{1+x^2}$

Substituting this value, we get

$\Rightarrow {\int }_1^1\frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)dx=-{\int }_{-1}^1\frac{1}{1+x^2}dx$

$\Rightarrow {\int }_1^1\frac{d}{dx}{\tan }^{-1}\left(\frac{1}{x}\right)dx=[-{\tan }^{-1}x{]}_{-1}^1$

$\Rightarrow -\frac{\pi }{4}-\frac{\pi }{4}=-\frac{\pi }{2}$