`int(sin(x-alpha))/(sin(x+alpha))dx=?`

Options

A. `xcos2alpha-sin2alpha*log|sin(x+alpha)|+C`
B. `xcos2alpha+sin2alpha*log|sin(x+alpha)|+C`
C. `xcos2alpha+sinalpha*log|sin(x+alpha)|+C`
D. none of these

Explanation

To solve the integral \(\int \frac{\sin(x - \alpha)}{\sin(x + \alpha)} \, dx\), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Rewrite the numerator**:
We can express \(\sin(x - \alpha)\) using the sine addition formula:
\[ \sin(x - \alpha) = \sin(x + \alpha - 2\alpha) = \sin(x + \alpha)\cos(2\alpha) - \cos(x + \alpha)\sin(2\alpha) \]
So, we rewrite the integral as:
\[ \int \frac{\sin(x + \alpha)\cos(2\alpha) - \cos(x + \alpha)\sin(2\alpha)}{\sin(x + \alpha)} \, dx \]

2. **Separate the integral**:
We can separate the integral into two parts:
\[ \int \left( \cos(2\alpha) - \frac{\cos(x + \alpha)\sin(2\alpha)}{\sin(x + \alpha)} \right) \, dx \]
This simplifies to:
\[ \int \cos(2\alpha) \, dx - \int \cot(x + \alpha) \sin(2\alpha) \, dx \]

3. **Integrate the first term**:
The integral of \(\cos(2\alpha)\) with respect to \(x\) is:
\[ \cos(2\alpha) \cdot x \]

4. **Integrate the second term**:
For the second term, we have:
\[ -\sin(2\alpha) \int \cot(x + \alpha) \, dx \]
The integral of \(\cot(x + \alpha)\) is:
\[ \ln|\sin(x + \alpha)| \]
Therefore, we have:
\[ -\sin(2\alpha) \ln|\sin(x + \alpha)| \]

5. **Combine the results**:
Putting it all together, we get:
\[ \int \frac{\sin(x - \alpha)}{\sin(x + \alpha)} \, dx = \cos(2\alpha) \cdot x - \sin(2\alpha) \ln|\sin(x + \alpha)| + C \]
where \(C\) is the constant of integration.

### Final Answer:
\[ \int \frac{\sin(x - \alpha)}{\sin(x + \alpha)} \, dx = \cos(2\alpha) \cdot x - \sin(2\alpha) \ln|\sin(x + \alpha)| + C \]