Let 2 - tan x 3 + tan x d x = 1 2 ( x + log e | sin x + cos x | ) + C , where C is the constant of integration. Then is equal to: [2024]

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Solution

Q.
Let $\int \frac{2 - \tan x}{3 + \tan x} d x = \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C,$ where $C$ is the constant of integration. Then $α + \frac{γ}{β}$ is equal to: [2024]

1 7

2 3

3 1

4 4

Ans.
(4)

Let $I = \int \frac{2 - \tan x}{3 + \tan x} d x = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} d x$

$Write Numerator = λ (Differentiation of Denominator) + μ (Denominator)$

$2 \cos x - \sin x = λ (- 3 \sin x + \cos x) + μ (3 \cos x + \sin x)$

$\Rightarrow λ + 3 μ = 2 \dots (i)$

$and - 3 λ + μ = - 1 \dots (ii)$

$From equations (i) and (ii), λ = μ = \frac{1}{2}$

$I = \frac{1}{2} \int \frac{- 3 \sin x + \cos x}{3 \cos x + \sin x} d x + \frac{1}{2} \int \frac{3 \cos x + \sin x}{3 \cos x + \sin x} d x$

$= \frac{1}{2} \int \frac{d t}{t} + \frac{1}{2} \int d x [Putting 3 \cos x + \sin x = t in first integral \Rightarrow (- 3 \sin x + \cos x) d x = d t]$

$= \frac{1}{2} \ln (t) + \frac{1}{2} x + C$

$= \frac{1}{2} \ln$ $| 3 \cos x + \sin x |$ $+ \frac{1}{2} x + C$

$= \frac{1}{2} (α x + \log_{e} | β \sin x + γ \cos x |) + C (∵Given)$

$\Rightarrow α = 1, β = 1, γ = 3$

$Now, α + \frac{γ}{β} = 1 + \frac{3}{1} = 4$

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