If `int(sinx)/(sin(x-alpha))dx=Ax+Blogsin(x-alpha)+c` then

Options

A. `A=sinalpha`
B. `B=sinalpha`
C. `A=cosalpha`
D. `B=cosalpha`

Answer

To solve the integral \( \int \frac{\sin x}{\sin(x - \alpha)} \, dx \) and express it in the form \( Ax + B \log \sin(x - \alpha) + C \), we can follow these steps:

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

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

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

### Step 4: Combine the Results
Putting it all together, we have:
\[ I = \cos \alpha \cdot x + \sin \alpha \log |\sin(x - \alpha)| + C \]

### Step 5: Identify A and B
From the expression \( I = Ax + B \log \sin(x - \alpha) + C \), we can identify:
- \( A = \cos \alpha \)
- \( B = \sin \alpha \)

### Final Result
Thus, we conclude:
\[ A = \cos \alpha, \quad B = \sin \alpha \]

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