The value of ∫01 log⁡sin⁡πx2dx is equal to

The value of 01logsinπx2dx is equal to

  1. A

    log 2

  2. B

    -log 2

  3. C

    log 3

  4. D

    0

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    Solution:

    Let, 

    I=01logsinπx2dxput, πx2=tdx=2πdt

    I=2π0π/2logsintdt=2πI1-----i

    where,

    I1=0π/2logsintdt=0π/2logsinπ2tdt=0π/2logcostdt

     2I1=0π/2(logsint+logcost)dt=0π/2log(sintcost)dt=0π/2logsin2t2dt=0π/2(logsin2tlog2)dt=120πlogsinzdzlog20π/2dt=1220π/2logsinzdzπ2log2 2I1=I1π2log2I1=π2log2

    From Eq. (i), I=2ππ2log2=log2

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