If $\int \frac{\sin x}{\sin (x-\alpha)} d x =A x+B \log \sin (x-\alpha)+C$, then the value of $(A, B)$, is
Answer & explanation
Correct answer: option 2
We have,
$\int \frac{\sin x}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha+\alpha)}{\sin (x-\alpha)} d x$
$=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} d x$
$=\int \cos \alpha d x+\sin \alpha \int \cot (x-\alpha) d x$
$=x \cos \alpha+\sin \alpha \log |\sin (x-\alpha)|+C$
∴ $A=\cos \alpha, B=\sin \alpha$
$\Rightarrow (A, B)=(\cos \alpha, \sin \alpha)$