Verifing ∫0πxln⁡(sin⁡x)dx=−ln⁡(2)π22

Dayton Burnett

Dayton Burnett

Answered

2022-01-23

Verifing 0πxln(sinx)dx=ln(2)π22

Answer & Explanation

pacetfv

pacetfv

Expert

2022-01-24Added 9 answers

Let I=0πxln(sinx)dx. Then, changing variables xπx:
I=0π(πx):
Therefore:
I=π20πln(sinx)dxsymmetry=π0π2ln(sinx)dx
The latter integral had been solved elsewhere.
Frauental91

Frauental91

Expert

2022-01-25Added 15 answers

This answer is meant to offer an alternative from the last integral of Sashas
RizerMix

RizerMix

Expert

2022-01-27Added 437 answers

For what is worth, the general case of what Sasha did is If I=0πxf(sinx)dx then I=π20πf(sinx)dx and the argument is completely analogous.

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