Evaluate the following Integrals :
`int (dx)/(sin x(3+cos x))`

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{dx}{\sin x (3 + \cos x)} \]
To simplify this, we can multiply and divide by \(\sin x\):
\[ I = \int \frac{\sin x \, dx}{\sin^2 x (3 + \cos x)} \]

### Step 2: Use the Identity for \(\sin^2 x\)
Recall the identity \(\sin^2 x = 1 - \cos^2 x\). Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin x \, dx}{(1 - \cos^2 x)(3 + \cos x)} \]

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

### Step 4: Partial Fraction Decomposition
Next, we will perform partial fraction decomposition on the integrand:
\[ \frac{1}{(1 - t^2)(3 + t)} = \frac{A}{1 - t} + \frac{B}{1 + t} + \frac{C}{3 + t} \]
Multiplying through by the denominator \((1 - t^2)(3 + t)\) and equating coefficients will help us find \(A\), \(B\), and \(C\).

### Step 5: Solve for Coefficients
Setting up the equation:
\[ 1 = A(1 + t)(3 + t) + B(1 - t)(3 + t) + C(1 - t^2) \]
By substituting convenient values for \(t\) (like \(t = 1\), \(t = -1\), and \(t = -3\)), we can solve for \(A\), \(B\), and \(C\).

1. **Let \( t = -1 \)**:
\[ 1 = A(0) + B(0) + C(4) \implies C = \frac{1}{4} \]

2. **Let \( t = 1 \)**:
\[ 1 = A(4) + B(0) + C(0) \implies A = \frac{1}{4} \]

3. **Let \( t = -3 \)**:
\[ 1 = A(0) + B(-4) + C(0) \implies B = -\frac{1}{4} \]

Thus, we have:
\[ A = \frac{1}{4}, \quad B = -\frac{1}{4}, \quad C = \frac{1}{4} \]

### Step 6: Rewrite the Integral
Substituting back into the integral:
\[ I = -\int \left( \frac{1/4}{1 - t} - \frac{1/4}{1 + t} + \frac{1/4}{3 + t} \right) dt \]
This simplifies to:
\[ I = -\frac{1}{4} \left( \int \frac{dt}{1 - t} - \int \frac{dt}{1 + t} + \int \frac{dt}{3 + t} \right) \]

### Step 7: Integrate Each Term
Now, we can integrate each term:
\[ I = -\frac{1}{4} \left( -\ln |1 - t| + \ln |1 + t| + \frac{1}{3} \ln |3 + t| \right) + C \]

### Step 8: Substitute Back
Substituting back \(t = \cos x\):
\[ I = -\frac{1}{4} \left( -\ln |1 - \cos x| + \ln |1 + \cos x| + \frac{1}{3} \ln |3 + \cos x| \right) + C \]

### Final Answer
Thus, the final answer for the integral is:
\[ I = \frac{1}{4} \ln \left( \frac{1 + \cos x}{1 - \cos x} \right) + \frac{1}{12} \ln |3 + \cos x| + C \]