Answered: "Verifing ∫0πxln⁡(sin⁡x)dx=−ln⁡(2)π22"

Dayton Burnett

Answered question

2022-01-23

Verifing

\int_{0}^{π} x \ln (\sin x) d x = - \ln (2) \frac{π^{2}}{2}

pacetfv

Beginner2022-01-24Added 9 answers

Let

I = \int_{0}^{π} x \ln (\sin x) d x

. Then, changing variables

x \to π - x

I = \int_{0}^{π} (π - x) :

Therefore:

I = \frac{π}{2} \int_{0}^{π} \ln (\sin x) {d x}^{symmetry} = π \int_{0}^{\frac{π}{2}} \ln (\sin x) d x

The latter integral had been solved elsewhere.

Frauental91

Beginner2022-01-25Added 15 answers

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

RizerMix

Expert2022-01-27Added 656 answers

For what is worth, the general case of what Sasha did is If

I = \int_{0}^{π} x f (\sin x) d x

then

I = \frac{π}{2} \int_{0}^{π} f (\sin x) d x

and the argument is completely analogous.

Do you have a similar question?

Recalculate according to your conditions!

Didn't find what you were looking for?