Question

If $\int \frac{\sin x}{\sin (x-\alpha)} d x=A x+ B \log \sin (x-\alpha)+C$, then value of $( A , B )$ is

• A. $(-\cos \alpha, \sin \alpha)$
• B. $(\cos \alpha, \sin \alpha)$
• C. $(-\sin \alpha, \cos \alpha)$
• D. $(\sin \alpha, \cos \alpha)$

Answer 1

Answer: (B)

$\int \frac{\sin x}{\sin (x-\alpha)} d x$
$=\int \frac{\sin (x-\alpha+\alpha)}{\sin (x-\alpha)} d x$
$=\int \frac{\sin (x-\alpha) \cos \alpha+\cos (x-\alpha) \sin \alpha}{\sin (x-\alpha)} d x$
$=\int\{\cos \alpha+\sin \alpha \cot (x-\alpha)\} d x$
$=(\cos \alpha) x+(\sin \alpha) \log \sin (x-\alpha)+C$
$A=\cos \alpha, B=\sin \alpha$