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JEE Main & Advanced Mathematics Definite Integration Question Bank Summation of series by Definite Integration, Gamma function, Leibnitz's rule

  • question_answer
    \[\int_{\,0}^{\,1}{\frac{d}{dx}\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]\,dx}\] is equal to  [Kerala (Engg.) 2002]

    A)                 0             

    B)                 \[\pi \]

    C)                 \[\pi /2\]             

    D)                 \[\pi /4\]

    Correct Answer: C

    Solution :

                    \[I=\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]_{0}^{1}={{\sin }^{-1}}(1)-{{\sin }^{-1}}(0)\]\[=\frac{\pi }{2}\].


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