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If $\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}=\mathrm{A} x+\mathrm{B} \log \sin (x-\alpha)+\mathrm{C}$, then value of $(A, B)$ is
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The correct answer is:
$(\cos \alpha, \sin \alpha)$
$\int \frac{\sin x}{\sin (x-\alpha)} \mathrm{dx}=\int \frac{\sin (x-\alpha+\alpha}{\sin (x-\alpha)} \mathrm{dx}$
$=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}$
$=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} \mathrm{d} \mathrm{x}$
$=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+\mathrm{C}$
$\therefore \mathrm{A}=\cos \alpha, \mathrm{B}=\sin \alpha$
$=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} \mathrm{d} \mathrm{x}$
$=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} \mathrm{d} \mathrm{x}$
$=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+\mathrm{C}$
$\therefore \mathrm{A}=\cos \alpha, \mathrm{B}=\sin \alpha$
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