If $\int \frac{\sin x}{\sin(x - \alpha)}dx = Ax + B\log\sin(x - \alpha) + c$, then
Step-by-Step Solution
Key Concept: Decompose $\sin x$ using the angle addition formula with $(x-\alpha)$ to separate the integral into a constant term and a logarithmic term.
We use the identity $\sin x = \sin(x - \alpha + \alpha) = \sin(x-\alpha)\cos\alpha + \cos(x-\alpha)\sin\alpha$. Therefore, $\frac{\sin x}{\sin(x-\alpha)} = \cos\alpha + \sin\alpha\cdot\frac{\cos(x-\alpha)}{\sin(x-\alpha)}$. Integrating both sides: $\int\frac{\sin x}{\sin(x-\alpha)}dx = \cos\alpha\cdot x + \sin\alpha\cdot\log|\sin(x-\alpha)| + c$. Comparing with $Ax + B\log\sin(x-\alpha) + c$, we get $A = \cos\alpha$ and $B = \sin\alpha$.
Correct Answer: 2,3