- #1

Tony1

- 17

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$$\int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx=-5\phi \pi$$

$\phi$ is the golden ratio

Make $u=\tan x$

$${1\over 2}\int_{0}^{\infty}u^{-{4\over 5}}\ln\left({1+u^2\over u^2}\right)\mathrm du$$

hmmm, too complicate to continue.
Any help, please. Thank you!

$\phi$ is the golden ratio

$${1\over 2}\int_{0}^{\infty}u^{-{4\over 5}}\ln\left({1+u^2\over u^2}\right)\mathrm du$$

hmmm, too complicate to continue.

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