Evaluate: `int(sinx)/(sin(x-a))\ dx`

Answer

To evaluate the integral \( I = \int \frac{\sin x}{\sin(x-a)} \, dx \), we can follow these steps:

### Step 1: Rewrite the Numerator
We can express the numerator \(\sin x\) in terms of \(\sin(x-a)\):
\[ \sin x = \sin((x-a) + a) = \sin(x-a)\cos a + \cos(x-a)\sin a \]
Thus, we rewrite the integral as:
\[ I = \int \frac{\sin((x-a) + a)}{\sin(x-a)} \, dx = \int \frac{\sin(x-a)\cos a + \cos(x-a)\sin a}{\sin(x-a)} \, dx \]

### Step 2: Split the Integral
Now, we can split the integral into two parts:
\[ I = \int \cos a \, dx + \int \frac{\cos(x-a)\sin a}{\sin(x-a)} \, dx \]
This simplifies to:
\[ I = \cos a \int dx + \sin a \int \cot(x-a) \, dx \]

### Step 3: Integrate Each Part
1. The first integral:
\[ \int dx = x \]
So, the first part becomes:
\[ \cos a \cdot x \]

2. The second integral:
Using the formula for the integral of \(\cot\):
\[ \int \cot u \, du = \log|\sin u| + C \]
Here, \(u = x - a\), thus:
\[ \int \cot(x-a) \, dx = \log|\sin(x-a)| + C \]
So, the second part becomes:
\[ \sin a \log|\sin(x-a)| \]

### Step 4: Combine the Results
Combining both parts, we get:
\[ I = \cos a \cdot x + \sin a \log|\sin(x-a)| + C \]

### Final Answer
Thus, the final result for the integral is:
\[ \int \frac{\sin x}{\sin(x-a)} \, dx = x \cos a + \sin a \log|\sin(x-a)| + C \]