Evaluating $\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\frac{G\ln 2}{2}$

Using Simpsons rule we know $$\cos(x)-\sin(x)=\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)$$ and $$\cos(x)+\sin(x)=\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)$$

Plugging this in the original integrals yields: $$\int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\sin\left(\frac{\pi}{4}-x\right)\right)\ln\left(\cos(x)\right)\mathrm{d}x- \int_0^{\frac{\pi}{4}}\ln\left(\sqrt{2}\cos\left(\frac{\pi}{4}-x\right)\right)\ln\left(\sin(x)\right)\mathrm{d}x$$

Splitting the logarithms and rearranging yields: $$\int_0^{\frac{\pi}{4}}\ln(\sqrt{2})\left(\ln(\cos(x))-\ln(\sin(x)\right)\mathrm{d}x+\\ +\left(\int_0^{\frac{\pi}{4}}\ln\left(\sin\left(\frac{\pi}{4}-x \right)\right)\ln(\cos(x)) \mathrm{d}x-\int_0^{\frac{\pi}{4}}\ln\left(\cos\left(\frac{\pi}{4}-x \right)\right)\ln(\sin(x)) \mathrm{d}x \right)$$

The first part yields $$\int_0^{\frac{\pi}{4}}\ln(\sqrt{2})\left(\ln(\cos(x))-\ln(\sin(x)\right)\mathrm{d}x = \ln(\sqrt{2})G=\frac{G\ln(2)}{2}$$

and because $$\int_0^bf(x)\mathrm{d}x=\int_0^bf(b-x)\mathrm{d}x$$ the second part yields $$\int_0^{\frac{\pi}{4}}\ln\left(\sin\left(\frac{\pi}{4}-x \right)\right)\ln(\cos(x)) \mathrm{d}x-\int_0^{\frac{\pi}{4}}\ln\left(\cos\left(\frac{\pi}{4}-x \right)\right)\ln(\sin(x)) \mathrm{d}x=0$$