Let ∫ 2 - tan x / 3 + tan x dx = 1 / 2 (αx + loge |β sin x + γ cos x |) + C, where C is the constant of integration.

Question

\(\text{Let}\,\int \frac{2-\tan x}{3+\tan x} d x=\frac{1}{2}\left(\alpha x+\log _{e}|\beta \sin x+\gamma \cos x|\right)+C\), where C is the constant of integration.

Then \(\alpha+\frac{\gamma}{\beta}\) is equal to : 

(1) 3

(2) 1

(3) 4

(4) 7

Answer 1

Correct option is : (3) 4 

\( \int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x\)

\(2 \cos x-\sin \mathrm{x}=\mathrm{A}(3 \cos \mathrm{x}+\sin \mathrm{x})+\mathrm{B}(\cos \mathrm{x}-3 \sin \mathrm{x})\)

\(3 \mathrm{A}+\mathrm{B}=2\)

A - 3B = - 1 

\(\Rightarrow \mathrm{A}=\frac{1}{2}, \mathrm{~B}=\frac{1}{2}\)

\(\therefore\ \int \frac{2 \cos \mathrm{x}-\sin \mathrm{x}}{3 \cos \mathrm{x}+\sin \mathrm{x}} \mathrm{dx}\)

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

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

\(=\frac{1}{2}(\alpha x+\ln |\beta \sin x+\gamma \cos x|)+C\)

\(\alpha=1, \beta=1, \gamma=3\)

\(\therefore \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4\)