2. Questions Bank
  3. JEE Main & Advanced
  4. Mathematics
  5. Definite Integration

JEE Main & Advanced Mathematics Definite Integration Question Bank Fundamental definite integration, Definite integration by substitution

  • question_answer
    The value of \[\int_{\,0}^{\,1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}\] is                               [RPET 2001]

    A)                 \[\pi /4\]             

    B)                 \[{{\pi }^{2}}/32\]

    C)                 1             

    D)                 None of these

    Correct Answer: B

    Solution :

               \[I=\int_{0}^{1}{\frac{{{\tan }^{-1}}x}{1+{{x}^{2}}}dx}\];  Put \[{{\tan }^{-1}}x=t\] Þ \[\frac{1}{1+{{x}^{2}}}dx=dt\]                                 \[\therefore I=\int_{\,0}^{\,\pi /4}{t\,dt}\]\[=\left[ \frac{{{t}^{2}}}{2} \right]_{0}^{\pi /4}\]\[=\frac{{{\pi }^{2}}}{32}\].


You need to login to perform this action.
You will be redirected in 3 sec spinner