A) \[(x-\alpha )\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c\]
B) \[(x-\alpha )\cos x+\log \sin (x-\alpha )+c\]
C) \[\sin (x-\alpha )+\sin x+c\]
D) \[\cos (x-\alpha )+\cos x+c\]
E) none of the above
Correct Answer: A
Solution :
Let\[I=\int{\frac{\sin x}{\sin (x-\alpha )}}dx\] Let \[x-\alpha =t\Rightarrow dx=dt\] \[\therefore \] \[I=\int{\frac{\sin (t+\alpha )}{\sin t}}dt\] \[=\int{\left( \frac{\sin t\cos \alpha }{\sin t}+\frac{\cos t\sin \alpha }{\sin t} \right)}dt\] \[=\cos \alpha \int{1\,dt}+\sin \alpha \int{\frac{\cos t}{\sin t}}dt\] \[=t\cos \alpha +\sin \alpha \log \sin t+c\] \[=(x-\alpha )\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c\]
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