Proving elementary, $\int_0^{2\pi}\log \frac{(1+\sin x)^{1+\cos x}}{1+\cos x} \mathrm{d}x=0$ The equation
$$
\int_{0}^{2\pi}\log\left(%
\left[1 + \sin\left(x\right)\right]^{1 + \cos\left(x\right)}
\over
1 + \cos\left(x\right)
\right)\,{\rm d}x = 0
$$.
Has been bothering me for a few days now. Note that the case from $0$ to $\pi/2$
has already been dealt with earlier. Some of the problems with the integral is the singularities at $x=\pi$ and $x=3\pi/2$, hower as the right and left limits agree the integral does not diverge.

Now I want to obtain the answer through substitutions and algebraic manipulation. I want to avoid all series representations (Catalan’s Constant) and alike.
I think one has to split the integrals to avoid the singularities and then show they are
alike without explicitly evalutating them. Note
$$
\int_0^{2\pi} f(x) \mathrm{d}x
=
\int_0^\pi f(x) \mathrm{d}x
+
\int_\pi^{3\pi/2} f(x) \mathrm{d}x
+
\int_{3\pi/2}^{2\pi} f(x) \mathrm{d}x
$$
Explicit calculations one have
$$
\int_0^{\pi/2} f(x)\mathrm{d}x = -1 + 2\log 2 \quad \, \quad
\int_{\pi}^{3\pi/2} f(x)\mathrm{d}x = 1 \\
\int_{\pi/2}^{\pi} f(x)\mathrm{d}x = 1 + 4 K-2 \log 2 \quad \, \quad
\int_{3\pi/2}^{2\pi} f(x)\mathrm{d}x = -1 - 4 K
$$
Where $K$ is Catalan's constant. Now the sum is of course zero, but is it possible to show this without evaluating $4$ integrals? The two first integrals I have been able to show
$$
\begin{align*}
        \int_{\pi}^{3\pi/2} f(x)\,\mathrm{d}x 
    & = \frac{1}{2}\int_{\pi}^{3\pi/2} f(x) + f(\pi-x)\,\mathrm{d}x \\
    & = \frac{1}{2}\int_{\pi}^{3\pi/2} \log(1+\cos x)\sin x+ \log(1+\sin x)\cos x\,\mathrm{d}x \\
    & = \frac{1}{2}\int_0^{-1} \log(1+u) + \log(1+u)\,\mathrm{d}x
      = 1
\end{align*}
$$
Is it perhaps possible to find
$$
\int_{3\pi/2}^{2\pi} f(x)\mathrm{d}x + 
\int_{\pi/2}^{\pi} f(x)\mathrm{d}x
= -2 \log 2
$$
?
 A: The last can be found elementarily without problems.
$$\begin{align}
I &= \int_{\pi/2}^\pi f(x)\,dx\\
&= \int_{\pi/2}^\pi (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx\tag{$x = y+\pi/2$}\\
&= \int_0^{\pi/2} (1-\sin y)\log (1+\cos y) - \log (1-\sin y)\,dy\\
II &= \int_{3\pi/2}^{2\pi} f(x)\,dx\\
&= \int_{3\pi/2}^{2\pi} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx \tag{$y = 2\pi-x$}\\
&= \int_0^{\pi/2} (1+\cos y)\log (1-\sin y) - \log (1+\cos y)\,dy\\
I+II &= \int_0^{\pi/2} \cos y \log (1-\sin y) - \sin y\log (1+\cos y)\,dy\\
&= \int_0^{\pi/2} \cos y \log (1-\sin y)\,dy\\
&\qquad + \int_0^{\pi/2} (-\sin y)\log (1+\cos y)\,dy\\
&= \int_0^1 \log (1-u)\,du + \int_1^0 \log (1+v)\,dv\\
&= \int_0^1 \log t\,dt - \int_1^2 \log t\,dt\\
&= \left[t\log t-t\right]_0^1 - \left[t\log t-t\right]_1^2\\
&= -1 - (2\log 2 - 2) + (-1)\\
&= -2\log 2.
\end{align}$$
If we further use
$$\begin{align}
III &= \int_\pi^{3\pi/2} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx\\
&= \int_0^{\pi/2} (1-\cos y)\log (1-\sin y) - \log (1-\cos y)\,dy\\
I+II+III &= \int_0^{\pi/2} \log (1-\sin y) - \log (1-\cos y) - \sin y\log (1+\cos y)\,dy,
\end{align}$$
the first two summands cancel (use $z = \pi/2-y$ on one of them), leaving
$$I+II+III = -\int_0^{\pi/2} \sin y \log (1+\cos y)\,dy = -\int_0^{\pi/2}\cos x \log (1+\sin x)\,dx.$$
Adding that to
$$\int_0^{\pi/2} (1+\cos x)\log (1+\sin x) - \log (1+\cos x)\,dx$$
we see that the entire integral is
$$\int_0^{\pi/2} \log (1+\sin x) - \log (1+\cos x)\,dx$$
which again cancels by symmetry, and we found
$$\int_0^{2\pi} \log \frac{(1+\sin x)^{1+\cos x}}{1+\cos x}\,dx = 0$$
by symmetry without evaluating a single integral explicitly.
A: $$
\begin{align}
&\int_0^{2\pi}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x\tag{1}\\
&=\int_0^{2\pi}\Big[(1+\cos(x))\log(1+\sin(x))-\log(1+\cos(x))\Big]\,\mathrm{d}x\tag{2}
\end{align}
$$
Since $\cos(\pi-x)=-\cos(x)$ and $\sin(\pi-x)=\sin(x)$, we get
$$
\begin{align}
&\int_0^{2\pi}(1+\cos(x))\log(1+\sin(x))\,\mathrm{d}x\tag{3}\\
&=\int_0^{2\pi}(1-\cos(x))\log(1+\sin(x))\,\mathrm{d}x\tag{4}\\
&=\int_0^{2\pi}\log(1+\sin(x))\,\mathrm{d}x\tag{5}\\
&=\int_0^{2\pi}\log(1+\cos(x))\,\mathrm{d}x\tag{6}
\end{align}
$$
Explanation:
$(4)$: substitute $x\mapsto\pi-x$
$(5)$: average $(3)$ and $(4)$
$(6)$: substitute $x\mapsto\pi/2-x$
Plugging $(6)$ into $(2)$, we get
$$
\int_0^{2\pi}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x=0\tag{7}
$$
