Recent content by Tony1

Prove that, $$\int_{0}^{\infty}{\tanh^2(t)\tanh(2t)\over t^2}\mathrm dt=\color{blue}{2\ln 2}$$

Note $$\sin(\pi/10)={1\over 2\phi}$$ $$-{5\over 2}\cdot {\pi\over \sin(\pi/10)}=-5\pi\phi$$

How may we go about to show that, $$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\color{green}{1\over \pi}\cdot\color{blue}{{\ln 2\over 2}(1-\ln 2)}$$

Hi greg1313, No, I have no solution for them, that why I am posting them for a solution.

How to prove this integral, $$\int_{0}^{2\pi}\sin\left({x\over 2}\right)\ln^2\left[\sin\left({x\over 4}\right)\sin\left({x\over 8}\right)\right]={27\over 2}+\ln^2(2)+\ln(2)$$

How to prove this integral, $$\int_{0}^{2\pi}\mathrm dt{\sin t\over \sin t+ i\sqrt{n+\cos^2 t}}={\pi\over 1+n}$$ $n \ne -1$ $i=\sqrt{-1}$

Given: A so-called complicate integral has a such a simple closed form, quite amazed me, but how to prove it, is an other story. $$\int_{0}^{1}\mathrm dt{t-3t^3+t^5\over 1+t^4+t^8}\cdot \ln(-\ln t) dt=\color{red}{{\pi\over 3\sqrt{3}}}\cdot \color{blue}{\ln 2\over 2}$$ Does anyone know to how...

Proposed: How can we prove $(1)?$ $$\int_{0}^{\infty}\mathrm dx{\sin^2\left({a\over x}\right)\over (4a^2+x^2)^2}={\pi\over (4a)^3}\tag1$$

$$\int_{0}^{\pi\over 2}{\ln(\sin^2 x)\over \sin(2x)}\cdot \sqrt[5]{\tan(x)}\mathrm dx=-5\phi \pi$$ $\phi$ is the golden ratio Any help, please. Thank you!