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JEE Main & Advanced Sample Paper JEE Main Sample Paper-11

  • question_answer
    \[\int\limits_{0}^{\pi /2}{x\left| {{\sin }^{2}}x-\frac{1}{2} \right|dx}\]is equal to

    A)  \[3\pi /4\]                         

    B)  \[\pi /4\]

    C)  \[\pi /8\]                            

    D)  \[\pi /2\]

    Correct Answer: C

    Solution :

     \[I=\frac{1}{2}\int\limits_{0}^{\pi /2}{x|\cos 2x|dx;2x=t\Rightarrow dx=\frac{dt}{2}}\] \[I=\frac{1}{8}\int\limits_{0}^{\pi }{t|\cos t|dt}\]             \[I=\frac{1}{8}\int\limits_{0}^{\pi }{(\pi -t)|\cot |dt}\] \[2I=\frac{1}{8}\int\limits_{0}^{\pi }{|\cos t|dt}=\frac{2\pi }{8}\Rightarrow I=\frac{\pi }{8}\]


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