Evaluate the following integrals
Evaluate `int (3 sin x+2 cos x)/(3 cos x + 2 sin x)dx`

Answer

To evaluate the integral

\[ I = \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx, \]

we can use a method involving the comparison of coefficients. Here’s a step-by-step solution:

### Step 1: Set up the integral
Let

\[ I = \int \frac{3 \sin x + 2 \cos x}{3 \cos x + 2 \sin x} \, dx. \]

### Step 2: Express the numerator in terms of the denominator
We can express the numerator as a linear combination of the denominator and its derivative. We write:

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

where \(A\) and \(B\) are constants we need to determine.

### Step 3: Differentiate the denominator
The derivative of the denominator \(3 \cos x + 2 \sin x\) is:

\[ \frac{d}{dx}(3 \cos x + 2 \sin x) = -3 \sin x + 2 \cos x. \]

### Step 4: Compare coefficients
Now we compare coefficients from both sides. Expanding the right-hand side gives:

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

From this, we can set up the following equations:

1. \(3A + 2B = 2\) (coefficient of \(\cos x\))
2. \(2A - 3B = 3\) (coefficient of \(\sin x\))

### Step 5: Solve the system of equations
We can solve these equations simultaneously. From the first equation:

\[ 3A + 2B = 2 \quad \text{(1)} \]

From the second equation:

\[ 2A - 3B = 3 \quad \text{(2)} \]

Now, we can multiply equation (1) by 3:

\[ 9A + 6B = 6 \quad \text{(3)} \]

And multiply equation (2) by 2:

\[ 4A - 6B = 6 \quad \text{(4)} \]

Now, adding equations (3) and (4):

\[ (9A + 6B) + (4A - 6B) = 6 + 6, \]
\[ 13A = 12 \implies A = \frac{12}{13}. \]

Substituting \(A\) back into equation (1):

\[ 3\left(\frac{12}{13}\right) + 2B = 2, \]
\[ \frac{36}{13} + 2B = 2 \implies 2B = 2 - \frac{36}{13} = \frac{26 - 36}{13} = \frac{-10}{13} \implies B = \frac{-5}{13}. \]

### Step 6: Substitute back into the integral
Now substitute \(A\) and \(B\) back into the integral:

\[ I = \int \left(\frac{12}{13} \cdot \frac{3 \cos x + 2 \sin x}{3 \cos x + 2 \sin x} + \frac{-5}{13} \cdot \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x}\right) \, dx. \]

This simplifies to:

\[ I = \frac{12}{13} \int dx - \frac{5}{13} \int \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x} \, dx. \]

### Step 7: Solve the remaining integral
Let \(t = 3 \cos x + 2 \sin x\). Then,

\[ dt = (-3 \sin x + 2 \cos x) \, dx. \]

Thus, we can rewrite the integral:

\[ \int \frac{2 \cos x - 3 \sin x}{3 \cos x + 2 \sin x} \, dx = \int \frac{dt}{t}. \]

### Step 8: Final integration
Now we can integrate:

\[ I = \frac{12}{13} x - \frac{5}{13} \ln |t| + C, \]

Substituting back \(t\):

\[ I = \frac{12}{13} x - \frac{5}{13} \ln |3 \cos x + 2 \sin x| + C. \]

### Final Answer
Thus, the final result is:

\[ I = \frac{12}{13} x - \frac{5}{13} \ln |3 \cos x + 2 \sin x| + C. \]