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NDA 2012
Exam
NDA
Year
2012
Subject
Mathematics
Chapter
Definite Integration

NDA 2012 · Mathematics · Definite Integration

What is ∫_0^1 ^ -1 x 1+x^2 d x equal to?

What is ∫01tan⁡−1x1+x2dx\int_0^1 \frac{\tan ^{-1} x}{1+x^2} d x equal to?

Options

  1. Aπ4\frac{\pi}{4}
  2. Bπ8\frac{\pi}{8}
  3. Cπ28\frac{\pi^2}{8}
  4. Dπ232\frac{\pi^2}{32}

Correct answer

D. π232\frac{\pi^2}{32}

Step-by-step solution

∫01tan⁡−1x1+x2dxPut tan⁡−1x=t11+x2dx=dtx=0,⇒t=0π/4∫0π/4tdtx=1,⇒t=π/4t22∫0π/412[(π4)2−(0)2]12(π16)2=π232.\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}
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