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

Explanation

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

### Step 1: Rewrite the Integral
We start by rewriting the integral:
\[ \int \frac{\sin x}{\sin(x-a)} \, dx = \int \frac{\sin(x-a + a)}{\sin(x-a)} \, dx \]
This allows us to use the sine addition formula.

### Step 2: Apply the Sine Addition Formula
Using the sine addition formula \(\sin(A + B) = \sin A \cos B + \cos A \sin B\), we can express \(\sin(x-a + a)\) as:
\[ \sin(x-a + a) = \sin(x-a) \cos a + \cos(x-a) \sin a \]
Substituting this back into the integral gives:
\[ \int \frac{\sin(x-a) \cos a + \cos(x-a) \sin a}{\sin(x-a)} \, dx \]

### Step 3: Simplify the Integral
This simplifies to:
\[ \int \left( \cos a + \frac{\cos(x-a) \sin a}{\sin(x-a)} \right) \, dx \]
Now we can separate the integral:
\[ \int \cos a \, dx + \int \frac{\cos(x-a) \sin a}{\sin(x-a)} \, dx \]

### Step 4: Evaluate the First Integral
The first integral is straightforward:
\[ \int \cos a \, dx = \cos a \cdot x \]

### Step 5: Evaluate the Second Integral
For the second integral, we can use the substitution:
Let \(t = \sin(x-a)\). Then, differentiating gives:
\[ dt = \cos(x-a) \, dx \quad \Rightarrow \quad dx = \frac{dt}{\cos(x-a)} \]
Now substituting \(t\) into the integral:
\[ \int \frac{\sin a}{t} \, dt \]
This integral evaluates to:
\[ \sin a \ln |t| + C = \sin a \ln |\sin(x-a)| + C \]

### Step 6: Combine Results
Combining both parts, we have:
\[ \int \frac{\sin x}{\sin(x-a)} \, dx = \cos a \cdot x + \sin a \ln |\sin(x-a)| + C \]

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