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NDA 2021
Exam
NDA
Year
2021
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2021 · Mathematics · Indefinite Integration

What is ∫ (d x)/( x+ x) equal to ?

What is ∫dxsec⁡x+tan⁡x\int \frac{d x}{\sec x+\tan x} equal to ?

Options

  1. Aln⁡(sec⁡x)+ln⁡∣sec⁡x+tan⁡x∣+c\ln (\sec x)+\ln |\sec x+\tan x|+c
  2. Bln⁡(sec⁡x)−ln⁡∣sec⁡x+tan⁡x∣+c\ln (\sec x)-\ln |\sec x+\tan x|+c
  3. Csec⁡xtan⁡x−ln⁡∣sec⁡x−tan⁡x∣+c\sec x \tan x-\ln |\sec x-\tan x|+c
  4. Dln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+c\ln |\sec x+\tan x|-\ln |\sec x|+c

Correct answer

D. ln⁡∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+c\ln |\sec x+\tan x|-\ln |\sec x|+c

Step-by-step solution

I=∫dxsec⁡x+tan⁡x=∫(sec⁡x−tan⁡x)⋅dx(sec⁡2x−tan⁡2x)=∫sec⁡x⋅dx−∫tan⁡x⋅dx=ln∣(sec⁡x+tan⁡x)∣−(−nn⁡∣cos⁡x∣)+C=ln∣sec⁡x+tan⁡x∣−ln⁡∣cos⁡x∣−1+C=ln∣sec⁡x+tan⁡x∣−ln⁡∣sec⁡x∣+C\begin{aligned} I & =\int \frac{d x}{\sec x+\tan x}=\int \frac{(\sec x-\tan x) \cdot d x}{\left(\sec ^2 x-\tan ^2 x\right)} \\ & =\int \sec x \cdot d x-\int \tan x \cdot d x \\ & =l n|(\sec x+\tan x)|-(-\operatorname{nn}|\cos x|)+C \\ & =l n|\sec x+\tan x|-l \operatorname{n}|\cos x|^{-1}+C \\ & =l n|\sec x+\tan x|-l \operatorname{n}|\sec x|+C\end{aligned}
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