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

  • question_answer
    The value of \[\int_{0}^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}\,dx}\] is                   [RPET 1995]

    A)                 \[\pi /2\]             

    B)                 \[\pi /4\]

    C)                 \[\pi /3\]             

    D)                 \[\pi /6\]

    Correct Answer: B

    Solution :

               \[I=\int_{0}^{\pi /2}{\frac{\sin x}{1+{{\cos }^{2}}x}}dx\]            Put \[\cos x=t\]Þ \[-\sin x\,dx=dt\]                                 Then \[I=\int_{1}^{0}{\frac{-dt}{1+{{t}^{2}}}=\int_{0}^{1}{\frac{dt}{1+{{t}^{2}}}=[{{\tan }^{-1}}t]_{0}^{1}=\frac{\pi }{4}}}\].


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