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Mathematics · Indefinite Integration

JEE Main 2024 — 9 April, Shift 1 — Question 7

Let ∫2−tan⁡x3+tan⁡xdx=12(αx+log⁡e∣βsin⁡x+γcos⁡x∣)+C\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _{e}|\beta \sin x+\gamma \cos x|\right)+C , where C is the constant of integration.

Then α+γβ\alpha+\frac{\gamma}{\beta} is equal to :

  1. Option A:

    3

  2. Option B:

    1

  3. Option C:

    4

    Correct
  4. Option D:

    7

Answer: C

Step-by-step solution

∫2−tan⁡x3+tan⁡xdx=∫2cos⁡x−sin⁡x3cos⁡x+sin⁡xdx\quad \int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x

2cos⁡x−sin⁡x=A(3cos⁡x+sin⁡x)+B(cos⁡x−3sin⁡x)2 \cos x-\sin x=A(3 \cos x+\sin x)+B(\cos x-3 \sin x)

3 A+B=23 \mathrm{~A}+\mathrm{B}=2 A−3B=−1A-3 B=-1

⇒A=12, B=12\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{1}{2}

∴∫2cos⁡x−sin⁡x3cos⁡x+sin⁡xdx\therefore \int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x

=x2+12ln⁡∣3cos⁡x+sin⁡x∣+C=\frac{x}{2}+\frac{1}{2} \ln |3 \cos x+\sin x|+C

=12(x+ln⁡∣3cos⁡x+sin⁡x∣)+C=\frac{1}{2}(x+\ln |3 \cos x+\sin x|)+C

=12(αx+ln⁡∣βsin⁡x+γcos⁡x∣)+C=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C

α=1,β=1,γ=3\alpha=1, \beta=1, \gamma=3

∴α+γβ=1+31=4\therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Mathematics
Chapter
Indefinite Integration
Topic
Methods of Indefinite Integration
Let int 2-tan x/3+tan x d x=1/2 (α x+log e β sin x+γ cos x )+C … | JEE Main 2024 PYQ with Solution · DhiX AI