`int(sinx)/(sin(x-alpha))dx =`

Options

A. `x cos alpha - sin alpha log sin (x-alpha) +c `
B. `x cos alpha + sin alpha log sin (x-alpha) +c `
C. `x sin alpha - sin alpha log sin (x-alpha)+c`
D. none of these

Answer

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

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\sin(x - \alpha)} \, dx \]

### Step 2: Use the Sine Difference Formula
We can express \(\sin(x - \alpha)\) using the sine difference formula:
\[ \sin(x - \alpha) = \sin x \cos \alpha - \cos x \sin \alpha \]
Thus, we can rewrite the integral as:
\[ I = \int \frac{\sin x}{\sin x \cos \alpha - \cos x \sin \alpha} \, dx \]

### Step 3: Simplify the Integral
To simplify the integral, we can separate the terms in the denominator:
\[ I = \int \frac{\sin x}{\sin x \cos \alpha - \cos x \sin \alpha} \, dx = \int \left( \frac{1}{\cos \alpha} + \frac{\sin \alpha \cos x}{\sin x (\cos \alpha - \sin \alpha)} \right) \, dx \]

### Step 4: Separate the Integral
Now we can separate the integral into two parts:
\[ I = \frac{1}{\cos \alpha} \int dx + \sin \alpha \int \frac{\cos x}{\sin x} \, dx \]

### Step 5: Integrate Each Part
1. The first integral:
\[ \int dx = x \]
Thus,
\[ \frac{1}{\cos \alpha} \int dx = \frac{x}{\cos \alpha} \]

2. The second integral:
\[ \int \frac{\cos x}{\sin x} \, dx = \log |\sin x| + C \]
Therefore,
\[ \sin \alpha \int \frac{\cos x}{\sin x} \, dx = \sin \alpha \log |\sin x| \]

### Step 6: Combine the Results
Combining both parts, we get:
\[ I = \frac{x}{\cos \alpha} + \sin \alpha \log |\sin x| + C \]

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

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