JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Integration by Substitution

  • question_answer
    \[\int{\frac{\sin x\,\,dx}{3+4{{\cos }^{2}}x}=}\] [Karnataka CET 2000]

    A)            \[\log (3+4{{\cos }^{2}}x)+c\]

    B)            \[\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}\left( \frac{\cos x}{\sqrt{3}} \right)+c\] 

    C)            \[\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\cos x}{\sqrt{3}} \right)+c\]

    D)            \[\frac{1}{2\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\cos x}{\sqrt{3}} \right)+c\]

    Correct Answer: C

    Solution :

                       \[I=\int{\frac{\sin x}{3+4{{\cos }^{2}}x}dx}\]            Put \[\cos x=t\]Þ \[-\sin x\,\,dx=dt\]            \[\therefore \,\,I=\int{\frac{-dt}{3+4{{t}^{2}}}}\]\[=\int{\frac{{{\sec }^{2}}x}{1-{{\tan }^{2}}x}dx}\]            Þ \[I=-\frac{1}{4.\frac{\sqrt{3}}{2}}.{{\tan }^{-1}}\frac{t}{\left( \frac{\sqrt{3}}{2} \right)}+c=\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}\frac{2\,t}{\sqrt{3}}+c\]                    \[\Rightarrow I=\frac{-1}{2\sqrt{3}}{{\tan }^{-1}}\left( \frac{2\cos x}{\sqrt{3}} \right)+c\].


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