A) \[(sin\text{ }\alpha ,\text{ }cos\text{ }\alpha )\]
B) (b)\[(cos\text{ }\alpha ,\text{ }sin\text{ }\alpha \text{)}\]
C) \[(-\sin \text{ }\alpha ,\text{ }\cos \text{ }\alpha )\]
D) \[(-\cos \text{ }\alpha ,\text{ }\sin \text{ }\alpha )\]
Correct Answer: B
Solution :
Let \[l=\int{\frac{\sin x}{\sin (x-\alpha )}}dx\] Put \[x-\alpha =t\Rightarrow dx=dt\] \[\therefore \] \[l=\int_{{}}^{{}}{\frac{\sin (t+\alpha )}{\sin t}dt}\] \[l=\int{\frac{\sin t\cos \alpha +\cos t\sin \alpha }{\sin t}}dt\] \[l=\int{\cos \alpha dt+}\int{\sin \alpha \frac{\cos t}{\sin t}}dt\] \[l=\cos \alpha \int{1dt+}\sin \alpha \int{\frac{\cos t}{\sin t}}dt\] \[l=\cos \alpha (t)+\sin \alpha \log \sin t+{{C}_{1}}\] \[l=\cos \alpha (x-\alpha )+\sin \alpha \log \sin (x-\alpha )+{{C}_{1}}\] \[l=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )-\alpha \cos \alpha +{{C}_{1}}\] \[l=x\cos \alpha +\sin \alpha \log \sin (x-\alpha )+C\] \[(\because \,let\,C=-\alpha \cos \alpha +{{C}_{1}})\] 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 w.r.t.\[x,\]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 the coefficients of \[\sin x\] and \[\cos x\] both sides, we get \[A\cos \alpha +B\sin \alpha =1\] ...(i) and \[B\cos \alpha -A\sin \alpha =0\] ...(ii) On solving Eqs. (i) and (ii), we get \[A=cos\alpha ,\text{ }B=sin\alpha \]
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