`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 will proceed step by step.

### Step 1: Substitution
Let us make the substitution:
\[ x - \alpha = t \quad \Rightarrow \quad dx = dt \]
This changes our integral to:
\[ \int \frac{\sin(t + \alpha) + \cos(t + \alpha)}{\sin t} \, dt \]

### Step 2: Expand Sine and Cosine
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 3: Simplify the Integral
Combining the terms in the numerator:
\[ \int \frac{\sin t \cos \alpha + \cos t \sin \alpha + \cos t \cos \alpha - \sin t \sin \alpha}{\sin t} \, dt \]
This can be separated into two integrals:
\[ \int \left( \cos \alpha + \frac{\cos t (\sin \alpha + \cos \alpha)}{\sin t} \right) \, dt \]

### Step 4: Further Simplification
This becomes:
\[ \int \cos \alpha \, dt + \int \frac{\cos t (\sin \alpha + \cos \alpha)}{\sin t} \, dt \]
The first integral is straightforward:
\[ \cos \alpha \cdot t \]

### Step 5: Second Integral
The second integral can be simplified as:
\[ \int \frac{\cos t}{\sin t} (\sin \alpha + \cos \alpha) \, dt = (\sin \alpha + \cos \alpha) \int \cot t \, dt \]
The integral of \( \cot t \) is:
\[ \ln |\sin t| + C \]

### Step 6: Combine Results
Putting everything together, we have:
\[ \cos \alpha \cdot t + (\sin \alpha + \cos \alpha) \ln |\sin t| + C \]
Substituting back \( t = x - \alpha \):
\[ \cos \alpha (x - \alpha) + (\sin \alpha + \cos \alpha) \ln |\sin(x - \alpha)| + C \]

### Final Result
Thus, the final result of the integral is:
\[ \int \frac{\sin x + \cos x}{\sin(x - \alpha)} \, dx = \cos \alpha (x - \alpha) + (\sin \alpha + \cos \alpha) \ln |\sin(x - \alpha)| + C \]