Evaluating $\int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\frac{G\ln 2}{2}$ In order to compute, in an elementary way,
$\displaystyle \int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$
(see Evaluating $\int_0^1 \frac{x \arctan x \log \left( 1-x^2\right)}{1+x^2}dx$ )
i need to show, in a simple way, that:
$\displaystyle \int_0^{\tfrac{\pi}{4}}\ln(\cos x-\sin x)\ln(\cos x) dx - \int_0^{\tfrac{\pi}{4}}\ln(\cos x+\sin x)\ln(\sin x)dx=\dfrac{G\ln 2}{2}$
$G$ being the Catalan constant.
The reason of my interest for this question is, if i am right, that this formula permits to find out a relation between integrals:
$\displaystyle \int_0^1 \dfrac{\ln(1+x)\ln(1+x^2)}{1+x^2}dx$
$\displaystyle \int_0^1 \dfrac{\ln(1+x)^2}{1+x^2}dx$
$\displaystyle \int_0^{\tfrac{\pi}{4}}\big(\ln(\cos x)\big)^2 \dfrac{}{}dx$
and some constants.

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
$$\int_0^{\frac{\pi}{4}}\ln\left(\sin\left(\frac{\pi}{4}-x \right)\right)\ln(\cos(x)) \mathrm{d}x-\int_0^{\frac{\pi}{4}}\ln\left(\cos\left(\frac{\pi}{4}-x \right)\right)\ln(\sin(x)) \mathrm{d}x=0$$
