Question

Let $\int \frac{2 - \tan x}{3 + \tan x} \, dx = \frac{1}{2} \left( \alpha x + \log_e \lvert \beta \sin x + \gamma \cos x \rvert \right) + C$, where $C$ is the constant of integration.

• A. 3
• B. 1
• C. 4
• D. 7

Answer 1

Answer: (C)

4

Answer 2

$\int \frac{2 - \tan x}{3 + \tan x} \, dx = \int \frac{2 \cos x - \sin x}{3 \cos x + \sin x} \, dx$

Let:

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

Equating coefficients:

$3A + B = 2, \quad A - 3B = -1$

Solving these equations:

$A = \frac{1}{2}, \quad B = \frac{1}{2}$

Thus:

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

Simplify:

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

Rewriting in the given form:

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

From the equation:

$\alpha = 1, \quad \beta = 1, \quad \gamma = 3$

Finally:

$\alpha + \frac{\gamma}{\beta} = 1 + \frac{3}{1} = 4$

Ans: Option (3): 4.