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JEE Main & Advanced Mathematics Indefinite Integrals Question Bank Fundamental Integration

  • question_answer
    \[\int_{{}}^{{}}{{{\tan }^{-1}}\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}}\ dx=\]

    A)            \[2{{x}^{2}}+c\]

    B)            \[{{x}^{2}}+c\]

    C)            \[\frac{{{x}^{2}}}{2}+c\]

    D)            \[2x+c\]

    Correct Answer: C

    Solution :

               \[\int_{{}}^{{}}{{{\tan }^{-1}}\sqrt{\frac{1-\cos 2x}{1+\cos 2x}}\,dx=\int_{{}}^{{}}{{{\tan }^{-1}}\tan x\,dx}}\]                    \[=\int_{{}}^{{}}{x\,dx=\frac{{{x}^{2}}}{2}}+c.\]


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