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NDA 2018
Exam
NDA
Year
2018
Subject
Mathematics
Chapter
Indefinite Integration

NDA 2018 · Mathematics · Indefinite Integration

What is ∫ d x a^2 ^2 x+b^2 ^2 x equal to?

What is ∫dxa2sin⁡2x+b2cos⁡2x\int \frac{d x}{a^2 \sin ^2 \mathrm{x}+\mathrm{b}^2 \cos ^2 \mathrm{x}} equal to?

Options

  1. Ac+1abtan⁡−1(atan⁡x b)\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\mathrm{a} \tan x}{\mathrm{~b}}\right)
  2. Bc−1abtan⁡−1( btan⁡xa)\mathrm{c}-\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\mathrm{~b} \tan \mathrm{x}}{\mathrm{a}}\right)
  3. Cc+1abtan⁡−1( btan⁡xa)\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\mathrm{~b} \tan \mathrm{x}}{\mathrm{a}}\right)
  4. DNone of the above

Correct answer

A. c+1abtan⁡−1(atan⁡x b)\mathrm{c}+\frac{1}{\mathrm{ab}} \tan ^{-1}\left(\frac{\mathrm{a} \tan x}{\mathrm{~b}}\right)

Step-by-step solution

I=1a2∫sec⁡2xtan⁡2x+(ba)2=1a2×abtan⁡−1(atan⁡xb)+c=c+1abtan⁡−1(atan⁡xb)\begin{aligned} & \mathrm{I}=\frac{1}{a^2} \int \frac{\sec ^2 x}{\tan ^2 x+\left(\frac{b}{a}\right)^2} \\ & =\frac{1}{a^2} \times \frac{a}{b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)+c \\ & =c+\frac{1}{a b} \tan ^{-1}\left(\frac{a \tan x}{b}\right)\end{aligned}
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