How may we go about to show that,
$$\int_{0}^{1}t\cos(2t\pi)\tan(t\pi)\ln[\sin(t\pi)]\mathrm dt=\color{green}{1\over \pi}\cdot\color{blue}{{\ln 2\over 2}(1-\ln 2)}$$
How to show that,
$$\int_{0}^{1}t^2\sec(t\pi)[1+\sin(t\pi)-\sin^2(t\pi)]\ln[\sin(t\pi)]\mathrm dt=\left({\pi\over 2}\right)^2-{\color{blue}{2\ln(2)\over \pi^2}}+\color{red}{{\ln^2(2)\over 2\pi}}$$
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 may one show that,
$$\int_{0}^{4\pi}\sin\left({x\over 2}\right)\sin\left({x\over 4}\right)\ln^2\left[\sin\left({x\over 8}\right)\right]\mathrm dx={\pi\over 9}[12\ln (2)-1]$$
Given that,
How to show that,
$$\int_{0}^{\infty}{\cos(\pi x^2)\over {1\over 2}+\cosh\left({x\pi\over \sqrt{3}}\right)}\mathrm dx=\sin\left(\pi\over 12\right)$$
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...