∫0π/2 sin⁡2xlog⁡tan⁡xdx is equal to 

 0π/2sin2xlogtanxdx is equal to 

  1. A

    π

  2. B

    π2

  3. C

    0

  4. D

    1

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    Solution:

    Let I=0π/2sin2xlogtanxdx

    Then,

    I0π/2sin2π2xlogtanπ2xdx I=0π/2sin2xlogcotxdx

    Adding (i) and (ii), we get

    2I=0π/2sin2x(logtanx+logcotx)dx 2I=0π/2sin2x×log1dx2I=0I=0

     

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