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JEE Main & Advanced JEE Main Paper (Held on 10-4-2019 Morning)

  • question_answer
    The value of \[\int\limits_{0}^{2\pi }{\left[ \sin 2x(1+cos3x) \right]}dx,\]where [t] denotes the greatest integer function, is : [JEE Main 10-4-2019 Morning]

    A) \[-2\pi \]                                  

    B) \[\pi \]

    C) \[-\pi \]

    D) \[2\pi \]

    Correct Answer: C

    Solution :

    \[I=\int\limits_{0}^{2\pi }{\left[ \sin 2x\left( 1+\cos 3x \right) \right]}dx\] \[I=\int\limits_{0}^{\pi }{\left( \left[ \sin 2x+\sin 2x\cos 3x \right]+ \right.}\] \[\left. \left[ -\sin 2x-\sin 2x\cos 3x \right] \right)dx\]\[=\int\limits_{0}^{\pi }{-dx}=-\pi \]                


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