If `int (3sin x+2 cosx)/(3cosx+2sinx)dx=ax+blog_(e)|2 sinx+3 cosx|+c` then a. `a= -(12)/(13)` b. `b=(6)/(13)` c. `a=(12)/(13)` d. `b= -(15)/(39)`

Options

A. `a= -(12)/(13)`
B. `b=(6)/(13)`
C. `a=(12)/(13)`
D. `b= -(15)/(39)`

Explanation

To solve the integral

\[ \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx = ax + b \log |2 \sin x + 3 \cos x| + c, \]

we will differentiate both sides and compare coefficients to find the values of \(a\) and \(b\).

### Step 1: Differentiate both sides

Differentiating the left-hand side:

\[ \frac{d}{dx} \left( \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx \right) = \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x}. \]

Differentiating the right-hand side:

\[ \frac{d}{dx} \left( ax + b \log |2 \sin x + 3 \cos x| + c \right) = a + b \cdot \frac{1}{2 \sin x + 3 \cos x} \cdot (2 \cos x - 3 \sin x). \]

### Step 2: Set the derivatives equal

Setting the derivatives equal gives us:

\[ \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} = a + b \cdot \frac{2 \cos x - 3 \sin x}{2 \sin x + 3 \cos x}. \]

### Step 3: Cross-multiply to eliminate the fraction

Cross-multiplying yields:

\[ (3 \sin x + 2 \cos x)(2 \sin x + 3 \cos x) = (3 \cos x + 2 \sin x)(a(2 \sin x + 3 \cos x) + b(2 \cos x - 3 \sin x)). \]

### Step 4: Expand both sides

Expanding the left-hand side:

\[ 6 \sin^2 x + 9 \sin x \cos x + 4 \cos^2 x. \]

Expanding the right-hand side:

\[ (3 \cos x + 2 \sin x)(a(2 \sin x + 3 \cos x) + b(2 \cos x - 3 \sin x)). \]

### Step 5: Collect terms and compare coefficients

From the expansion, we will collect coefficients of \(\sin x\) and \(\cos x\) and set them equal to the corresponding coefficients from the left-hand side.

1. Coefficient of \(\cos x\):

\[ 3a + 2b = 2 \]

2. Coefficient of \(\sin x\):

\[ 2a - 3b = 3 \]

### Step 6: Solve the system of equations

We have the system of equations:

1. \(3a + 2b = 2\)
2. \(2a - 3b = 3\)

Multiply the first equation by 2:

\[ 6a + 4b = 4. \]

Multiply the second equation by 3:

\[ 6a - 9b = 9. \]

Subtract the first modified equation from the second:

\[ (6a - 9b) - (6a + 4b) = 9 - 4, \]
which simplifies to:

\[ -13b = 5 \implies b = -\frac{5}{13}. \]

### Step 7: Substitute \(b\) back to find \(a\)

Substituting \(b\) into the first equation:

\[ 3a + 2\left(-\frac{5}{13}\right) = 2, \]
which simplifies to:

\[ 3a - \frac{10}{13} = 2 \implies 3a = 2 + \frac{10}{13} = \frac{26}{13} + \frac{10}{13} = \frac{36}{13}. \]
Thus,

\[ a = \frac{12}{13}. \]

### Final Results

The values are:

- \(a = \frac{12}{13}\)
- \(b = -\frac{5}{13}\)

### Conclusion

The correct options are:
- \(a = \frac{12}{13}\) (Option C)
- \(b = -\frac{5}{13}\)