Answer:
Option B
Explanation:
$\int_{}^{}\frac{\sin x}{\sin (x - \alpha)}dx$
= $\int_{}^{}\frac{\sin (x-\alpha+\alpha)}{\sin (x - \alpha)}dx$
=$\int_{}^{}\frac{\sin (x-\alpha)\cos\alpha+\cos (x-\alpha)\sin\alpha}{\sin (x - \alpha)}dx$
=$\int_{}^{}\left\{\cos\alpha+\sin\alpha\cot(x - \alpha)\right\}dx$
= $(\cos\alpha)x+(\sin\alpha)\log_{}{\sin(x-\alpha)}+C$
A = $\cos\alpha$ B = $\sin\alpha$
