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JEE Main
Exam
JEE Main
Subject
Mathematics
Chapter
Indefinite Integration

JEE Main · Mathematics · Indefinite Integration

∫ e^ x [(2 + 2x)/(1 + 2x)]dx =

∫ex[2+sin⁡2x1+cos⁡2x]dx=∫ e^{x}\left[\dfrac{2 + \sin 2x}{1 + \cos 2x}\right]dx =

Options

  1. Aextanx+Ce^{x}tan x + C
  2. Bex+tanx+Ce^{x} + tan x + C
  3. C2extanx+C2e^{x}tan x + C
  4. Dextan⁡2x+Ce^{x}\tan 2x + C

Correct answer

A. extanx+Ce^{x}tan x + C

Step-by-step solution

Let,& I=∫ex[2+sin⁡2x1+cos⁡2x]dxI = ∫ e^{x}\left[\dfrac{2 + \sin 2x}{1 + \cos 2x}\right]dx

We know that~sin⁡2x=2sin⁡xcos⁡x and cos⁡2x=2cos⁡2x−1\sin 2x = 2\sin x \cos x \text{ and } \cos 2x = 2\cos^{2}x - 1

Using the above formulas we can write

I=∫ex(2+2sin⁡xcos⁡x2cos⁡2x)dxI = ∫ e^{x}\left(\dfrac{2 + 2\sin x \cos x}{2\cos^{2}x}\right)dx

I=∫ex(2(1+sin⁡xcos⁡x)2cos⁡2x)dxI = ∫ e^{x}\left(\dfrac{2\left(1 + \sin x \cos x\right)}{2\cos^{2}x}\right)dx

I=∫ex(1cos⁡2x+sin⁡xcos⁡xcos⁡2x)dxI = ∫ e^{x}\left(\dfrac{1}{\cos^{2}x} + \dfrac{\sin x \cos x}{\cos^{2}x}\right)dx
I=∫ex(tan⁡x+sec⁡2x)dxI = ∫ e^{x} ( \tan x + \sec^{2} x )dx

We know that~∫ex(f(x)+f′(x))dx=ex⋅f+C∫ e^{x}\left(\text{f}\left(\text{x}\right) + f^{'}\left(\text{x}\right)\right)\text{dx} = e^{x} \cdot \text{f} + \text{C}

Using the above formula we can write

I=extan⁡x+CI = e^{x} \tan x + C

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