Question

If $ \int{\frac{\sin x}{\sin (x-\alpha )}}dx=Ax+B\log |sin (x-\alpha )|+C, $ then the value of $ A-B $ at $ \alpha =\frac{\pi }{2} $ is

• A. $ -1 $
• B. $1$
• C. $2$
• D. $0$
• E. $ -2 $

Answer 1

Answer: (E)

Let $ I=\int{\frac{\sin x}{\sin (x-\alpha )}}dx $ Put $ x-\alpha =r\Rightarrow dx=dt $
$ \therefore $ $ I=\int{\frac{\sin (t+\alpha )}{\sin t}}dt $
$ \Rightarrow $ $ I=\int{\cos \alpha \,dt}+\int{\sin \alpha }\frac{\cos t}{\sin t}dt $
$ \Rightarrow $ $ I=\cos \alpha \int{1\,dt}+\sin \alpha \int{\frac{\cos t}{\sin t}}dt $
$ \Rightarrow $ $ I=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+{{C}_{1}} $
$ \Rightarrow $
$=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C $ But $ \int{\frac{\sin x}{\sin (x-\alpha )}}dx $
$=Ax+B\log \sin (x-\alpha )+C $
$ \therefore $ $ x\cos \alpha +\sin \alpha \log \sin (x-a)+C $
$=Ax+B\log \sin (x-\alpha )+C $ At $ \alpha =\frac{\pi }{2},A=0 $ and $ B=1 $
$ \therefore $ $ A-B=-1 $