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Question

Prove π20log(sinx)dx=π2log2

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Solution

I=π20log(sinx)dx

I=π20log(cosx)dx

2I=π20[log(sinx)+log(cosx)]dx

2I=π20log(sinxcosx)dx

2I=π20log(sin2x2)dx

2I=π20log(sin2x)dxπ20log2dx

2I=I1log2[x]π20

2I=I1π2log2...(1)

I1=π20log(sin2x)dx

substitute 2x=tdt=2dxx=0π2t=0π

I1=12π0log(sint)dt

I1=I...(2)

Placing (2) in (1)

2I=Iπ2log2

I=π2log2

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