How to calculate $\int_0^\pi \ln(1+\sin x)\mathrm dx$ How to calculate this integral
$$\int_0^\pi \ln(1+\sin x)\mathrm dx$$
I didn't find this question in the previous questions. With the help of Wolframalpha I got an answer $-\pi \ln 2+4\mathbf{G}$, where $\mathbf{G}$ denotes Catalan's Constant.
 A: First we rewrite $\log(1+\sin(x))=\log(2)+2\log(\sin(x/2+\pi/4))$. This yields, after shifting $x/2+\pi/4\rightarrow y,dx\rightarrow 2 dy$
$$
I=\pi\log(2)+4\underbrace{\int_{\pi/4}^{3\pi/4}\log(\sin(y))dy}_{J}
$$
Now we might employ the Fourier series of $\log(\sin(x))$ to calculate $J$
$$
J=-\log(2)\int_{\pi/4}^{3\pi/4}dy-\sum_{n=1}^{\infty}\frac{1}{k}\int_{\pi/4}^{3\pi/4}\cos(2yk)dy=\\
-\frac{\pi}{2}\log(2)+\sum_{k=1}^{\infty}\frac{(-1)^k}{(2k+1)^2}
$$
where we used $\sin(\frac{3\pi}{2} k)-\sin(\frac{\pi}{2} k)=(-1)^{k}-(-1)^{k+1}=2(-1)^k$ in the second line.
Employing  the series representation of the Catalan constant we find
$$
J=-\frac{\pi}{2}\log(2)+G
$$
and therefore

$$
I=\pi \log(2)+J=-\pi \log(2)+4G \quad (*)
$$


Just for fun, lets see what contour integration can do.
We again use $J$ rewriting it by the help of the identity $\log(\sin(x))=\log(i/2)-ix+\log(1-e^{2 i x})$. The integral over the first two terms is trivial leaving us with
$$
J=-\frac{\pi}{2}\log(2)+\underbrace{\int_{\pi/4}^{3\pi/4}\log(1-e^{2 i x})}_{K}
$$
to evaluate $K$ we integrate the complex valued function
$$
f(z)=\log(1-e^{2 i z})
$$ 
over an rectangle $C$ in the complex plane with verticies $(\pi/4,3\pi/4,3\pi/4+i R,\pi/4+i R)$ where in the end we want to take the limit  $R\rightarrow +\infty$. As one can easily check, the contribution from the top of the rectangle vanishs (the integrand vanishs as $\mathcal{O}(e^{-2R})$ for big $R$). Furthermore the integrand is holomorphic inside the contour of integration and therefore we can express the integral of interest in terms of the vertical pieces of the contour
$$
\int_C f(z)dz=K+i\int_{0}^{\infty}\log(1+ie^{-2y})dy-i\int_{0}^{\infty}\log(1-ie^{-2y})dy=0
$$
this can be simplified to
$$
K=2\int_0^{\infty}\arctan(e^{-2y})dy=\int_0^{\infty}\arctan(e^{-y})dy
$$
using the series repesentation of arctan, we may easily conclude that this integral is equal to $G$ and therefore

$$
K=G
$$

from which it follows that
$$
J=-\frac{\pi}{2}\log(2)+K=-\frac{\pi}{2}\log(2)+G$$
from which (*) follows immediatly 
