JEE Main & Advanced AIEEE Solved Paper-2004

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

    A)

    B)                        1             

    C)                        2             

    D)        3

    Correct Answer: C

    Solution :

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


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