A) \[(sin\alpha ,\,cos\alpha )\]
B) \[(cos\alpha ,\,\sin \alpha )\]
C) \[(-\sin \alpha ,\,\cos \alpha )\]
D) \[(-\cos \alpha ,\,\sin \alpha )\]
Correct Answer: B
Solution :
[b] \[I=\int{\frac{\sin x}{\sin (x-\alpha )}dx}\] Let \[x-\alpha =t\Rightarrow dx=dt\therefore x=(t+\alpha )\] \[\therefore I=\int{\frac{\sin (t+\alpha )}{\sin t}dt}\] \[=\int{\frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}dt}\] \[=\int{\cos \alpha \,dt+\int{\sin \alpha \frac{\cos t}{\sin t}dt}}\] \[=\cos \alpha \int{1dt+\sin \alpha \int{\frac{\cos t}{\sin t}dt}}\] \[=\cos \alpha (t)+sin\alpha \,log\,\sin t+c\] \[=\cos \alpha (x-\alpha )+\sin \alpha \log \,\sin (x-\alpha )+c\] \[=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )-\alpha cos\alpha +c\]\[=x\cos \alpha +sing\,\alpha log\,sin(x-\alpha )+c\] But \[\int{\frac{\sin x}{\sin (x-\alpha )}dx=Ax+B\,\,\log \sin (x-\alpha )+c}\] \[\therefore x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c\] \[=Ax+B\,\,\log \sin (x-\alpha )+c\] On comparing, we get \[A=\cos \alpha \] \[B=\sin \alpha \]
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