Q235 / 270 · NDA 2018 (Phase 2) · Mathematics

What is $\int_{-1}^1\left\{\frac{d}{d x}\left(\tan ^{-1} \frac{1}{x}\right)\right\} d x$ equal to?

  • A
    0
  • B
    $-\frac{\pi}{4}$
  • Correct answer
    C
    $-\frac{\pi}{2}$
  • D
    $\frac{\pi}{2}$

Solution

$\begin{aligned} & \int_{-1}^1\left(\frac{d}{d x} \tan ^{-1} \frac{1}{x}\right) d x \\ & =\int_{-1}^1\left(\frac{d}{d x} \cot ^{-1} x\right) d x \\ & =\int_{-1}^1 \frac{-1}{\left(1+x^2\right)} d x=-2 \int_0^1 \frac{1}{1+x^2} d x \\ & =-2\left[\tan ^{-1} x\right]_0^1=-2 \times \frac{\pi}{4}=-\frac{\pi}{2}\end{aligned}$

Hence, the correct answer is option: 'C'.

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