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

  • question_answer
    \[\int_{{}}^{{}}{\frac{\sin 2xdx}{1+{{\cos }^{2}}x}}=\]      [Karnataka CET 2005]

    A)                 \[\frac{1}{2}\log (1+{{\cos }^{2}}x)+c\]

    B)                 \[2\log (1+{{\cos }^{2}}x)+c\]

    C)                 \[\frac{1}{2}\log (1+\cos 2x)+c\]

    D)                 \[-\log (1+{{\cos }^{2}}x)+c\]

    Correct Answer: D

    Solution :

                       \[I=\int{\frac{\sin 2x}{1+{{\cos }^{2}}x}dx=\int{\frac{2\sin x\cos x}{1+{{\cos }^{2}}x}dx}}\]            Put \[1+{{\cos }^{2}}x=t\] Þ \[-2\sin x\cos x\,\,dx=dt\]            Þ \[\sin 2x=-dt\]. Hence \[I=\int{^{-}\left( \frac{dt}{t} \right)}=-\log t+c\]                 \[=-\log (1+{{\cos }^{2}}x)+c\].


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