JEE Main & Advanced AIEEE Solved Paper-2004

  • question_answer
    If \[\int{\frac{\sin x}{\sin (x-\alpha )}}dx\] \[=Ax+B\text{ }\log \text{ }\sin (x-\alpha )+C,\]then value of \[(A,\text{ }B)\]is

    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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