A) \[(\sin \alpha ,\cos \alpha )\]
B) \[(\cos \alpha ,\sin \alpha )\]
C) \[(-\sin \alpha ,\cos \alpha )\]
D) \[(-\cos \alpha ,\sin \alpha )\]
Correct Answer: B
Solution :
Let \[I=\int{\frac{\sin x}{\sin (x-\alpha )}}dx\] Put \[x-\alpha =t\Rightarrow dx=dt\] \[I=\int{\frac{\sin (t+\alpha )}{\sin t}}dt\] \[I=\int{\frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}}dt\] \[I=\int{\cos \alpha \,dt}+\int{\sin \alpha \frac{\cos t}{\sin t}}dt\] \[I=\cos \alpha \int{1dt}+\sin \alpha \int{\frac{\cos t}{\sin t}}dt\] \[I=\cos \alpha (t)+\sin \alpha \log \sin t+{{c}_{1}}\] \[I=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+{{c}_{1}}\] \[I=\cos \alpha +\sin \alpha \log \sin (x-\alpha )-\alpha \cos \alpha +{{c}_{1}}\] \[I=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c\] But \[\int{\frac{\sin x}{\sin (x-\alpha )}}dx=Ax+B\log \sin (x-\alpha )+c\] \[\therefore \] \[x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+c\] \[=Ax+B\log \sin (x-\alpha )+c\] On comparing, we get \[A=cos\text{ }\alpha ,\text{ }B=sin\text{ }\alpha \] Alternate Solution \[\because \] \[\int{\frac{\sin x}{\sin (x-\alpha )}}dx=Ax+B\log \sin (x-\alpha )+c\] On differentiating both sides with respect to\[x\] we get \[\frac{\sin x}{\sin (x-\alpha )}=A+B\frac{\cos (x-\alpha )}{\sin (x-\alpha )}\] \[\Rightarrow \] \[\sin x=A\sin (x-\alpha )+B\cos (x-\alpha )\] \[\Rightarrow \] \[\sin x=A(\sin x\cos \alpha -\cos x\sin \alpha )\] \[+B(\cos x\cos \alpha +\sin x\sin \alpha )\] \[\Rightarrow \] \[\sin x=\sin x(A\cos \alpha +B\sin \alpha )\] \[+\cos x(B\cos \alpha -A\sin \alpha )\] On comparing, we get \[A\text{ }cos\,\alpha +B\text{ }sin\,\alpha =1\] ...(i) and \[B\text{ }cos\,\alpha -A\text{ }sin\,\alpha =0\] ... (ii) On solving Eqs. (i) and (ii), we get \[A=cos\,\alpha ,\text{ }B=sin\,\alpha \]
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