`int (1)/(sin x(3+2 cos x))dx`

Answer

To solve the integral \( I = \int \frac{1}{\sin x (3 + 2 \cos x)} \, dx \), we will follow a systematic approach.

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ I = \int \frac{1}{\sin x (3 + 2 \cos x)} \, dx \]

### Step 2: Multiply by \(\sin x\)
We can multiply and divide by \(\sin x\):
\[ I = \int \frac{\sin x}{\sin^2 x (3 + 2 \cos x)} \, dx = \int \frac{\sin x}{(1 - \cos^2 x)(3 + 2 \cos x)} \, dx \]

### Step 3: Substitution
Let \( t = \cos x \). Then, \( dt = -\sin x \, dx \) or \( \sin x \, dx = -dt \). Substituting these into the integral gives:
\[ I = -\int \frac{dt}{(1 - t^2)(3 + 2t)} \]

### Step 4: Factor the Denominator
We can rewrite the denominator:
\[ 1 - t^2 = (1 - t)(1 + t) \]
Thus, the integral becomes:
\[ I = -\int \frac{dt}{(1 - t)(1 + t)(3 + 2t)} \]

### Step 5: Partial Fraction Decomposition
We will perform partial fraction decomposition:
\[ \frac{1}{(1 - t)(1 + t)(3 + 2t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{3 + 2t} \]
Multiplying through by the denominator:
\[ 1 = A(1 + t)(3 + 2t) + B(1 - t)(3 + 2t) + C(1 - t)(1 + t) \]

### Step 6: Solve for Coefficients
Expanding and equating coefficients will give us a system of equations to solve for \( A \), \( B \), and \( C \).

1. Set \( t = 1 \):
\[ 1 = A(2)(5) \implies A = \frac{1}{10} \]

2. Set \( t = -1 \):
\[ 1 = B(2)(1) \implies B = \frac{1}{2} \]

3. Set \( t = -\frac{3}{2} \):
\[ 1 = C\left(-\frac{1}{2}\right)\left(\frac{1}{2}\right) \implies C = \frac{4}{5} \]

### Step 7: Substitute Back into the Integral
Now substituting back, we have:
\[ I = -\left( \frac{1}{10} \int \frac{dt}{1 - t} + \frac{1}{2} \int \frac{dt}{1 + t} + \frac{4}{5} \int \frac{dt}{3 + 2t} \right) \]

### Step 8: Integrate Each Term
Integrating each term:
1. \(\int \frac{dt}{1 - t} = -\log |1 - t|\)
2. \(\int \frac{dt}{1 + t} = \log |1 + t|\)
3. \(\int \frac{dt}{3 + 2t} = \frac{1}{2} \log |3 + 2t|\)

Thus:
\[ I = -\left( \frac{1}{10} (-\log |1 - t|) + \frac{1}{2} \log |1 + t| + \frac{4}{5} \cdot \frac{1}{2} \log |3 + 2t| \right) + C \]

### Step 9: Substitute Back for \( t \)
Substituting back \( t = \cos x \):
\[ I = \frac{1}{10} \log |1 - \cos x| - \frac{1}{2} \log |1 + \cos x| - \frac{2}{5} \log |3 + 2 \cos x| + C \]

### Final Answer
Thus, the final answer is:
\[ I = \frac{1}{10} \log |1 - \cos x| - \frac{1}{2} \log |1 + \cos x| - \frac{2}{5} \log |3 + 2 \cos x| + C \]