Evaluate $\int_{\frac{-\pi}4}^{\frac{\pi}4}\ln(\sin x+\cos x)\mathrm{d}x$ $$\int_{\frac{-\pi}4}^{\frac{\pi}4} \ln(\sin x+\cos x)\mathrm{d}x $$
I just can't think of any technique to solve this question. 
Can anyone help me with at least how to begin?

Suppose one knows that,
1)$\displaystyle \int_0^1 \dfrac{\ln x}{1+x^2}dx=-G$
2) $\displaystyle \int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-G$
Where $G$ is the Catalan constant.
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\sin x+\cos x)dx$
$I=\displaystyle \int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln\left(\dfrac{\sin x+\cos x}{\cos x}\right)dx+\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\cos x)dx$
Since $x\rightarrow \ln(\cos x)$ is an even function,
$\displaystyle I=\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\tan x+1)dx+2\int_{0}^{\tfrac{\pi}{4}}\ln(\cos x)dx$
$\displaystyle I=\int_{-\tfrac{\pi}{4}}^{\tfrac{\pi}{4}}\ln(\tan x+1)dx-\int_{0}^{\tfrac{\pi}{4}}\ln((\sec x)^2)dx$
Perform the change of variable $y=\tan x$,
$\displaystyle I=\int_{-1}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
$\displaystyle I=\int_{-1}^0 \dfrac{\ln(1+x)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Perform the change of variable $y=-x$ in the first integral.
$\displaystyle I=\int_{0}^1 \dfrac{\ln(1-x)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$ in the first integral:
$\displaystyle I=\int_{0}^1 \dfrac{\ln\left(\dfrac{2x}{1+x}\right)}{1+x^2}dx+\int_{0}^1 \dfrac{\ln(1+x)}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx$
Therefore,
$\displaystyle I=\dfrac{\pi\ln 2}{4}+\int_0^1 \dfrac{\ln x}{1+x^2}dx-\int_{0}^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{4}-G+G-\dfrac{\pi\ln 2}{2}=-\dfrac{\pi\ln 2}{4}$
