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

  • question_answer
    \[\int{\frac{dx}{\sin x(3+{{\cos }^{2}}x)}}\] is equal to

    A) \[\log \left| {{y}^{2}}-1 \right|-{{\tan }^{-1}}y+C\]

    B) \[{{\tan }^{-1}}\frac{y}{\sqrt{3}}+C\]

    C) \[\log \left| \frac{y-1}{y+1} \right|+C\]

    D) \[\frac{1}{4}\log \left| \frac{y-1}{y+1} \right|-\frac{1}{4\sqrt{3}}{{\tan }^{-1}}\frac{y}{\sqrt{3}}+C\]

    Correct Answer: D

    Solution :

    [d] \[\int{\frac{dx}{\sin x(3+{{\cos }^{2}}x)}}\] \[=\int{\frac{\sin x\,dx}{{{\sin }^{2}}x(3+{{\cos }^{2}}x)}}\] \[=\int{\frac{\sin x\,dx}{(1-{{\cos }^{2}}x)(3+{{\cos }^{2}}x)}}\] \[=\int{\frac{dy}{({{y}^{2}}-1)({{y}^{2}}+3)}}\] (Putting \[\cos \,\,x=y\]) \[=\frac{1}{4}\int{\left[ \frac{1}{{{y}^{2}}-1}-\frac{1}{{{y}^{2}}+3} \right]dy}\] \[=\frac{1}{4}\log \left| \frac{y-1}{y+1} \right|-\frac{1}{4\sqrt{3}}{{\tan }^{-1}}\frac{y}{\sqrt{3}}+C\]


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