Q239 / 270 · NDA 2012 (Phase 1) · Mathematics

What is $\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x$ equal to?

  • A
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{8}$
  • C
    $\frac{\pi^2}{8}$
  • Correct answer
    D
    $\frac{\pi^2}{32}$

Solution

$\begin{aligned} & \int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x \\ & \text {Put } \tan ^{-1} x=t \\ & \frac{1}{1+x^2} d x=d t \\ & x=0, \Rightarrow t=0 \\ & \pi / 4 \\ & \int_0^{\pi / 4} t d t \quad x=1, \Rightarrow t=\pi / 4 \\ & \frac{t^2}{2} \int_0^{\pi / 4} \\ & \frac{1}{2}\left[\left(\frac{\pi}{4}\right)^2-(0)^2\right] \quad \frac{1}{2}\left(\frac{\pi}{16}\right)^2=\frac{\pi^2}{32} .\end{aligned}$

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

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