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JEE Main & Advanced Sample Paper JEE Main Sample Paper-33

  • question_answer
    The value of \[\int\limits_{0}^{\pi /2}{\left( \frac{1+\sin 3x}{1+2\sin x} \right)}dx,\], is equal to

    A)  \[\frac{2}{3}\]          

    B)                                     \[\frac{\pi }{4}\]

    C)  1                 

    D)                    \[\frac{3\pi }{4}\]

    Correct Answer: D

    Solution :

    \[\int\limits_{0}^{\pi /2}{\left( \frac{1+\sin 3x}{1+2\sin x} \right)\,dx=\int\limits_{0}^{\pi /2}{\left( \frac{1+3\sin x\,-4{{\sin }^{3}}x}{1+2\sin x} \right)}}\]             = 1


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