If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+c$, then the value of $\mathrm{A}$…
If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+B \log \sin (x-\alpha)+c$, then the value of $\mathrm{A}$ and $\mathrm{B}$ are respectively (where $\mathrm{c}$ is a constant of integration)
$\cos \alpha, \sin \alpha$
$\sin \alpha, \cos \alpha$
$-\cos \alpha, \sin \alpha$
$-\sin \alpha, \cos \alpha$
Solution
$\begin{aligned}
& \text { Let } I=\int \frac{\sin x}{\sin (x-\alpha)} d x \\
& =\int \frac{\sin [(x-\alpha)+\alpha]}{\sin (x-\alpha)} d x=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \\
& =\cos \alpha \int d x+\sin \alpha \int \frac{\cos (x-\alpha)}{\sin (x-\alpha)} d x=(\cos \alpha)(x)+(\sin \alpha) \\
& \log |\sin (x-\alpha)|+c
\end{aligned}$
Comparing with given data, we get $\mathrm{A}=\cos \alpha$ and $\mathrm{B}=\sin \alpha$