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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 1

  • question_answer
    \[\int\limits_{0}^{1}{\frac{d}{dx}}\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]dx\]is equal to

    A)  0                    

    B)  \[\pi \]    

    C)  \[\pi /2\]           

    D) \[\pi /4\]

    Correct Answer: C

    Solution :

    [c]:\[\int\limits_{0}^{1}{\frac{d}{dx}}\left[ {{\sin }^{-1}}\left( \frac{2x}{1+{{x}^{2}}} \right) \right]dx\] \[=\int\limits_{0}^{1}{\frac{d}{dx}}(2ta{{n}^{-1}}x)dx\] \[=\int\limits_{0}^{1}{\frac{2}{1+{{x}^{2}}}}dx=2\left[ {{\tan }^{-1}}x \right]_{0}^{1}=2\left( \frac{\pi }{4} \right)=\frac{\pi }{2}\]


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