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

  • question_answer
    The value of \[I=\int_{0}^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{1+\sin 2x}}\text{ }}dx\]is    [AIEEE 2004]

    A)                 3             

    B)                 1

    C)                 2             

    D)                 0

    Correct Answer: C

    Solution :

               \[I=\int_{0}^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{1+\sin 2x}}dx}\]\[=\int_{0}^{\pi /2}{\frac{{{(\sin x+\cos x)}^{2}}}{\sqrt{{{(\sin x+\cos x)}^{2}}}}dx}\]                    \[I=\int_{0}^{\pi /2}{(\sin x+\cos x)dx=(-\cos x+\sin x)_{0}^{\pi /2}}\]                                 \[I=1-(-1)=2\].


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