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Question: The value of \(\int_{- 1}^{1}{\frac{d}{dx}\left( \tan^{- 1}\frac{1}{x} \right)dx}\) is...

The value of ∫−11ddx(tan⁡−11x)dx\int_{- 1}^{1}{\frac{d}{dx}\left( \tan^{- 1}\frac{1}{x} \right)dx} is

A

π/2

B

π/4

C

-π/2

D

None of these

Answer

-π/2

Explanation

Solution

We have,

ddx(tan⁡−11x)=ddx(cot⁡−1x)=−11+x2\frac{d}{dx}\left( \tan^{- 1}\frac{1}{x} \right) = \frac{d}{dx}\left( \cot^{- 1}x \right) = - \frac{1}{1 + x^{2}}

∴\therefore ∫−11ddx(tan⁡−11x)dx=∫−11−11+x2dx=−2∫0111+x2dx\int_{- 1}^{1}{\frac{d}{dx}\left( \tan^{- 1}\frac{1}{x} \right)dx = \int_{- 1}^{1}{\frac{- 1}{1 + x^{2}}dx = - 2\int_{0}^{1}{\frac{1}{1 + x^{2}}dx}}}

=−2[tan⁡−1x]01=−2(π4)=−π2- 2\left\lbrack \tan^{- 1}x \right\rbrack_{0}^{1} = - 2\left( \frac{\pi}{4} \right) = - \frac{\pi}{2}