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NDA 2017
Exam
NDA
Year
2017
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2017 · Mathematics · Indefinite Integration

What is ∫ ^ -1 ( x+ x) d x equal to?

What is ∫tan⁡−1(sec⁡x+tan⁡x)dx\int \tan ^{-1}(\sec x+\tan x) d x equal to?

Options

  1. Aπx4+x24+c\frac{\pi x}{4}+\frac{x^2}{4}+c
  2. Bπx2+x24+c\frac{\pi \mathrm{x}}{2}+\frac{\mathrm{x}^2}{4}+\mathrm{c}
  3. Cπx4+πx24+c\frac{\pi \mathrm{x}}{4}+\frac{\pi \mathrm{x}^2}{4}+\mathrm{c}
  4. Dπx4−x24+c\frac{\pi x}{4}-\frac{x^2}{4}+c

Correct answer

A. πx4+x24+c\frac{\pi x}{4}+\frac{x^2}{4}+c

Step-by-step solution

∫tan⁡−1(sec⁡x+tan⁡x)dx=∫tan⁡−1(tan⁡(π4+x2))dx∵sec⁡x+tan⁡x=1+sin⁡xcos⁡x\begin{aligned} & \int \tan ^{-1}(\sec x+\tan x) d x \\ & =\int \tan ^{-1}\left(\tan \left(\frac{\pi}{4}+\frac{x}{2}\right)\right) d x \quad \because \sec x+\tan x=\frac{1+\sin x}{\cos x}\end{aligned}=∫(π4+x2)dx=πx4+x24+c\begin{aligned} & =\int\left(\frac{\pi}{4}+\frac{x}{2}\right) d x \\ & =\frac{\pi x}{4}+\frac{x^2}{4}+c\end{aligned}$
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