JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Self Evaluation Test - Integrals

  • question_answer
    The value of \[\int{\frac{\sin x}{\sin 4x}dx}\] is

    A) \[\frac{1}{4}\log \left| \frac{\sin x-1}{\sin x+1} \right|-\frac{1}{\sqrt{2}}\log \left| \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} \right|+C\]

    B) \[\frac{1}{8}\log \left| \frac{\cos x-1}{\cos x+1} \right|-\frac{1}{2\sqrt{2}}\log \left| \frac{\sqrt{2}\cos x-1}{\sqrt{2}\cos x+1} \right|+C\]

    C) \[\frac{1}{8}\log \left| \frac{\sin x-1}{sinx+1} \right|-\frac{1}{4\sqrt{2}}\log \left| \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} \right|+C\]

    D) None of these.

    Correct Answer: C

    Solution :

    [c] \[I=\int{\frac{\sin \,\,x\,\,dx}{4\sin x\cos x\cos 2x}}\] \[=\frac{1}{4}\int{\frac{\cos x\,dx}{(1-{{\sin }^{2}}x)(1-2\,{{\sin }^{2}}x)}}\] \[=\frac{1}{4}\int{\frac{dt}{(1-{{t}^{2}})(1-2{{t}^{2}})}}\] \[[t=sin\,x]\] \[=\frac{1}{4}\int{\left( \frac{2}{1-2{{t}^{2}}}-\frac{1}{1-{{t}^{2}}} \right)dt}\] \[=\frac{1}{4}\left\{ \frac{2}{2\sqrt{2}}\log \left| \frac{1+\sqrt{2}t}{1-\sqrt{2}t} \right|-\frac{1}{2}\log \left| \frac{1+t}{1-t} \right| \right\}+C\] \[=\frac{1}{8}\log \left| \frac{\sin x-1}{\sin x+1} \right|-\frac{1}{4\sqrt{2}}\log \left| \frac{\sqrt{2}\sin x-1}{\sqrt{2}\sin x+1} \right|+C\]


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