A) \[-1\]
B) 1
C) 2
D) 0
E) \[-2\]
Correct Answer: E
Solution :
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 \[B=1\] \[\therefore \] \[A-B=-1\]
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