If `int (sin x)/(sin (x-a))dx =Ax +B log sin (x-alpha ) +C,` then value of (A,B) is:

Options

A. `(sin alpha ,cos alpha )`
B. `(cos alpha, sin alpha )`
C. `(-sin alpha, cos alpha )`
D. `(-cos alpha, sin alpha )`

Answer

To solve the integral \( \int \frac{\sin x}{\sin(x - a)} \, dx \) and find the values of \( A \) and \( B \) in the expression \( Ax + B \log \sin(x - \alpha) + C \), we will follow these steps:

### Step 1: Rewrite the Integral
We start with the integral:
\[ I = \int \frac{\sin x}{\sin(x - a)} \, dx \]
We can express \( \sin x \) using the sine addition formula:
\[ \sin x = \sin((x - a) + a) = \sin(x - a) \cos a + \cos(x - a) \sin a \]
Thus, we rewrite the integral as:
\[ I = \int \frac{\sin(x - a) \cos a + \cos(x - a) \sin a}{\sin(x - a)} \, dx \]

### Step 2: Separate the Integral
Now, we can separate the integral into two parts:
\[ I = \int \cos a \, dx + \int \frac{\cos(x - a) \sin a}{\sin(x - a)} \, dx \]
The first integral is straightforward:
\[ \int \cos a \, dx = \cos a \cdot x \]

### Step 3: Simplify the Second Integral
For the second integral, we have:
\[ \int \frac{\cos(x - a) \sin a}{\sin(x - a)} \, dx = \sin a \int \cot(x - a) \, dx \]
The integral of \( \cot(x - a) \) is:
\[ \int \cot(x - a) \, dx = \log |\sin(x - a)| + C \]
Thus, we can write:
\[ \sin a \int \cot(x - a) \, dx = \sin a \log |\sin(x - a)| \]

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

### Step 5: Identify Coefficients
Now, we can compare this result with the given form \( Ax + B \log \sin(x - \alpha) + C \):
- From \( I = \cos a \cdot x + \sin a \log |\sin(x - a)| + C \), we identify:
- \( A = \cos a \)
- \( B = \sin a \)

### Final Result
Thus, the values of \( (A, B) \) are:
\[ (A, B) = (\cos a, \sin a) \]