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

  • question_answer
    \[\int{\frac{dx}{\cos x+\sqrt{3}\sin x}}\] equals

    A) \[\log \tan \left( \frac{x}{2}+\frac{\pi }{12} \right)+C\]

    B) \[\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\]

    C) \[\frac{1}{2}\log \tan \left( \frac{x}{2}+\frac{\pi }{12} \right)+C\]

    D) \[\frac{1}{2}\log \tan \left( \frac{x}{2}-\frac{\pi }{12} \right)+C\]

    Correct Answer: C

    Solution :

    [c] \[I=\int{\frac{dx}{\cos x+\sqrt{3}\sin x}}\] \[\Rightarrow I=\int{\frac{dx}{2\left[ \frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x \right]}}\] \[=\frac{1}{2}\int{\frac{dx}{\left[ \sin \frac{\pi }{6}\cos x+\cos \,\frac{\pi }{6}\sin x \right]}}\] \[=\frac{1}{2}.\int{\frac{dx}{\sin \left( x+\frac{\pi }{6} \right)}}\] \[\Rightarrow I=\frac{1}{2}.\int{\cos ec\left( x+\frac{\pi }{6} \right)dx}\] \[\because \int{\cos ec\,\,x\,\,dx=\log \left| (\tan x/2) \right|+C}\] \[\therefore I=\frac{1}{2}.\log \tan \left( \frac{x}{2}+\frac{\pi }{12} \right)+C\]


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