Prove that $\int_{0}^{2\pi}\ln \left(\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}\right)dx = 0$ 
Prove that
$$
I=\int_{0}^{2\pi}\ln \left(\frac{(1+\sin x)^{1+\cos x}}{1+\cos x}\right)\;dx = 0
$$

My Attempt:
$$
\begin{align}
I &= \int_{0}^{2\pi}\ln(1+\sin x)^{1+\cos x}\;dx-\int_{0}^{2\pi}\ln(1+\cos x)\;dx\\
&=\int_{0}^{2\pi}(1+\cos x)\cdot \ln(1+\sin x)\;dx-\int_{0}^{2\pi}\ln(1+\cos x)\;dx\\
&= \int_{0}^{2\pi}(1+\cos x)\cdot \ln(1+\sin x)\;dx - 2\int_{0}^{\pi}\ln(1+\cos x)\;dx
\end{align}
$$
How can I complete the solution from this point?
 A: $$
\begin{align}
&\int_0^{2\pi}\log\left(\frac{(1+\sin(x))^{1+\cos(x)}}{1+\cos(x)}\right)\,\mathrm{d}x\\
&=\int_0^{2\pi}\left[(1+\cos(x))\log(1+\sin(x))-\log(1+\cos(x))\right]\,\mathrm{d}x\tag{1}\\
&=\color{#C00000}{\int_0^{2\pi}\log(1+\sin(x))\,\mathrm{d}x-\int_0^{2\pi}\log(1+\cos(x))\,\mathrm{d}x}\\
&+\color{#00A000}{\int_0^{2\pi}\cos(x)\log(1+\sin(x))\,\mathrm{d}x}\tag{2}\\
&=\color{#C00000}{0}+\color{#00A000}{\int_0^{2\pi}\log(1+\sin(x))\,\mathrm{d}(1+\sin(x))}\tag{3}\\
&=\int_1^1\log(u)\,\mathrm{d}u\tag{4}\\[9pt]
&=0\tag{5}
\end{align}
$$
Explanation:
$(1)$: properties of $\log$
$(2)$: redistributing pieces
$(3)$: the red integrals cancel using $\cos(x)=\sin(x+\pi/2)$
$(4)$: substitute $u=1+\sin(x)$
$(5)$: integral that ends where it starts is $0$

Alternatively, note that
$$
\begin{align}
\cos(x)\log(1+\sin(x))
&=\cos(u+\pi/2)\log(1+\sin(u+\pi/2))\\
&=-\sin(u)\log(1+\cos(u))
\end{align}
$$
is an odd function of $u$. Therefore, because the integral of an odd function is $0$, we get by the substitution $x=u+\pi/2$,
$$
\begin{align}
\int_0^{2\pi}\cos(x)\log(1+\sin(x))\,\mathrm{d}x
&=-\int_0^{2\pi}\sin(u)\log(1+\cos(u))\,\mathrm{d}u\\
&=-\int_{-\pi}^\pi\sin(u)\log(1+\cos(u))\,\mathrm{d}u\\[6pt]
&=0
\end{align}
$$
A: Using the last step of your work, we have
$$
\int_0^{\large2\pi}\ln(1+\sin x)\ dx+\int_0^{\large2\pi}\ln(1+\sin x)\cdot\cos x\ dx-\int_0^{\large2\pi}\ln(1+\cos x)\ dx.
$$
By symmetry, we obtain
$$
\int_0^{\large2\pi}\ln(1+\sin x)\ dx=\int_0^{\large2\pi}\ln(1+\cos x)\ dx.
$$
Letting $y=1+\sin x$ the middle integral turns out to be
$$
\int_{x=0}^{\large2\pi}\ln(1+\sin x)\cdot\cos x\ dx=\int_{y=1}^{1}\ln y\ dy=0.
$$

Addendum :
Consider the plot of $\ln(1+\sin x)\cdot\cos x$ for $0<x<2\pi$,

then, by symmetry,
$$
\int_{x=0}^{\large2\pi}\ln(1+\sin x)\cdot\cos x\ dx=0.
$$
A: Let's make use of the general fact that, when integrating over an entire period, the sine and cosine functions are interchangeable, as are their signs, i.e., for any function of two variables, we have
$$\int_0^{2\pi}f(\cos\theta,\sin\theta)d\theta = \int_0^{2\pi}f(\sin\theta,\cos\theta)d\theta$$
and
$$\int_0^{2\pi}f(\cos\theta,\sin\theta)d\theta = \int_0^{2\pi}f(-\cos\theta,\sin\theta)d\theta=\int_0^{2\pi}f(\cos\theta,-\sin\theta)d\theta$$
Thus
$$I=\int_0^{2\pi}\log\left({(1+\sin\theta)^{1+\cos\theta}\over1+\cos\theta} \right)d\theta$$
implies
$$\begin{align}
2I&=\int_0^{2\pi}\left[\log\left({(1+\sin\theta)^{1+\cos\theta}\over1+\cos\theta} \right)+\log\left({(1+\sin\theta)^{1-\cos\theta}\over1-\cos\theta} \right)\right]d\theta\\
&=\int_0^{2\pi}\log\left({(1+\sin\theta)^2\over(1+\cos\theta)(1-\cos\theta)} \right)d\theta\\
&=2\int_0^{2\pi}\log(1+\sin\theta)d\theta-\int_0^{2\pi}\log(1+\cos\theta)d\theta-\int_0^{2\pi}\log(1-\cos\theta)d\theta\\
&=2\int_0^{2\pi}\log(1+\cos\theta)d\theta-\int_0^{2\pi}\log(1+\cos\theta)d\theta-\int_0^{2\pi}\log(1+\cos\theta)d\theta\\
&=0
\end{align}$$
