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

The Correct Option is C

We start with the given integral:

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

Now, express the numerator as a linear combination of the derivatives of the denominator terms:

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

Expanding and comparing coefficients, we get:

\( 3A + B = 2 \)

\( A - 3B = -1 \)

Solving these two equations:

\( A = \frac{1}{2}, \quad B = \frac{1}{2} \)

Hence,

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

Simplifying further:

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

This can be generalized as:

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

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

Therefore,

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

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.