`int(sin x+cos x)/(sin(x -alpha))dx` is equal to

Options

A. `(cos alpha- sin alpha)(x - alpha)+(cosalpha+sinalpha)log|sin(x-a)|+C`
B. `(cos alpha+sin alpha)(x - alpha)+(cosalpha+sinalpha)log|sin(x-a)|+C`
C. `(cos alpha+sin alpha)(x + alpha)+(cosalpha+sinalpha)log|sin(x+a)|+C`
D. none of these

Explanation

To solve the integral \(\int \frac{\sin x + \cos x}{\sin(x - \alpha)} \, dx\), we can follow these steps:

### Step 1: Substitute Variables
Let \(t = x - \alpha\). Then, \(dx = dt\) and \(x = t + \alpha\).

### Step 2: Rewrite the Integral
Substituting \(x\) in the integral, we get:
\[ \int \frac{\sin(t + \alpha) + \cos(t + \alpha)}{\sin t} \, dt \]

### Step 3: Use Angle Addition Formulas
Using the angle addition formulas:
\[ \sin(t + \alpha) = \sin t \cos \alpha + \cos t \sin \alpha \]
\[ \cos(t + \alpha) = \cos t \cos \alpha - \sin t \sin \alpha \]
Substituting these into the integral gives:
\[ \int \frac{(\sin t \cos \alpha + \cos t \sin \alpha) + (\cos t \cos \alpha - \sin t \sin \alpha)}{\sin t} \, dt \]

### Step 4: Simplify the Integral
Combining the terms in the numerator:
\[ \int \frac{(\sin t \cos \alpha - \sin t \sin \alpha) + (2 \cos t \cos \alpha)}{\sin t} \, dt \]
This simplifies to:
\[ \int \left(\cos \alpha - \sin \alpha + 2 \cot t \cos \alpha\right) \, dt \]

### Step 5: Separate the Integral
Now, we can separate the integral:
\[ \int (\cos \alpha - \sin \alpha) \, dt + 2 \cos \alpha \int \cot t \, dt \]

### Step 6: Integrate Each Term
1. The first integral:
\[ \int (\cos \alpha - \sin \alpha) \, dt = (\cos \alpha - \sin \alpha)t + C_1 \]
2. The second integral:
\[ \int \cot t \, dt = \log |\sin t| + C_2 \]
So, we have:
\[ 2 \cos \alpha \log |\sin t| + C_2 \]

### Step 7: Combine Results
Combining both parts, we get:
\[ (\cos \alpha - \sin \alpha)(t) + 2 \cos \alpha \log |\sin t| + C \]

### Step 8: Substitute Back for \(t\)
Now, substitute back \(t = x - \alpha\):
\[ (\cos \alpha - \sin \alpha)(x - \alpha) + 2 \cos \alpha \log |\sin(x - \alpha)| + C \]

### Final Answer
Thus, the integral evaluates to:
\[ \int \frac{\sin x + \cos x}{\sin(x - \alpha)} \, dx = (\cos \alpha - \sin \alpha)(x - \alpha) + 2 \cos \alpha \log |\sin(x - \alpha)| + C \]