Evaluating $\int_{0}^{\pi}\ln (1+\cos x)\, dx$ The problem is
$$\int_{0}^{\pi}\ln (1+\cos x)\ dx$$
What I tried was using standard limit formulas like changing $x$ to $\pi - x$ and I also tried integration by parts on it to no avail. Please help. Also this is my first question so please tell if I am wrong somewhere.
 A: Another way to solve is to use the Cauchy Integral Formula
$$ f(z)=\frac{1}{2\pi i}\int_{\partial D}\frac{f(\xi)}{\xi-z}d\xi$$
where $f(z)$ is analytic in $D$ and continuous in $\bar{D}$. Note that
$$ |1+\cos t+i\sin t|^2=2(1+\cos t).$$
So
\begin{eqnarray}
&&\int_{0}^\pi\log(1+\cos t)\; dt\\
&=&\frac12\int_{0}^{2\pi}\log(1+\cos t)\; dt\\
&=&\frac12\int_{0}^{2\pi}\log|1+\cos t+i\sin t|\; dt-\pi\log2\\
&=&\frac12\int_{0}^{2\pi}\Re\left[\log(1+\cos t+i\sin t)\right]\; dt-\pi\log2\\
&=&\frac12\Re\left[\int_{0}^{2\pi}\log(1+\cos t+i\sin t)\; dt\right]-\pi\log2\\
&=&\frac12\Re\left[\int_{|z|=1}\log(1+z)\; \frac{dz}{iz}\right]-\pi\log2\\
&=&\frac12\times2\pi \log 1-\pi\log2\\
&=&-\pi\log2.
\end{eqnarray}
A: Using Fourier Series
As shown in this answer,
$$
\log(1+\cos(x))=2\sum_{k=1}^\infty(-1)^{k-1}\frac{\cos(kx)}{k}-\log(2)\tag{1}
$$
For all $k\in\mathbb{Z}\setminus\{0\}$
$$
\int_0^\pi\cos(kx)\,\mathrm{d}x=0\tag{2}
$$
Therefore,
$$
\int_0^\pi\log(1+\cos(x))\,\mathrm{d}x=-\pi\log(2)\tag{3}
$$

A More Elementary Approach
$$
\begin{align}
\int_0^\pi\log(1+\cos(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(2\cos^2\left(\frac x2\right)\right)\,\mathrm{d}x\\
&=\pi\log(2)+\int_0^\pi\log\left(\cos^2\left(\frac x2\right)\right)\,\mathrm{d}x\\
&=\pi\log(2)+2\int_0^{\pi/2}\log\left(\cos^2(x)\right)\,\mathrm{d}x\tag{4}
\end{align}
$$
and
$$
\begin{align}
\int_0^\pi\log(1-\cos(x))\,\mathrm{d}x
&=\int_0^\pi\log\left(2\sin^2\left(\frac x2\right)\right)\,\mathrm{d}x\\
&=\pi\log(2)+\int_0^\pi\log\left(\sin^2\left(\frac x2\right)\right)\,\mathrm{d}x\\
&=\pi\log(2)+2\int_0^{\pi/2}\log\left(\sin^2(x)\right)\,\mathrm{d}x\tag{5}
\end{align}
$$
By substituting $x\mapsto\pi-x$, we see that the left side of $(4)$ equals the left side of $(5)$. Therefore,
$$
\begin{align}
2\int_0^\pi\log(1+\cos(x))\,\mathrm{d}x
&=\int_0^\pi\log(1+\cos(x))\,\mathrm{d}x+\int_0^\pi\log(1-\cos(x))\,\mathrm{d}x\tag{6}\\
&=\int_0^\pi\log\left(\sin^2(x)\right)\,\mathrm{d}x\tag{7}\\
&=2\int_0^\pi\log(1-\cos(x))\,\mathrm{d}x\tag{8}\\
&=2\pi\log(2)+4\int_0^{\pi/2}\log\left(\sin^2(x)\right)\,\mathrm{d}x\tag{9}\\
&=2\pi\log(2)+2\int_0^\pi\log\left(\sin^2(x)\right)\,\mathrm{d}x\tag{10}\\
&=-2\pi\log(2)\tag{11}
\end{align}
$$
Explanation:
$\phantom{1}(6)$: the left side of $(4)$ equals the left side of $(5)$
$\phantom{1}(7)$: add the integrands
$\phantom{1}(8)$: twice the left side of $(4)$ equals twice the left side of $(5)$
$\phantom{1}(9)$: apply $(5)$
$(10)$: $\sin(x)=\sin(\pi-x)$
$(11)$: $2$ times $(7)$ minus $(10)$
Thus, dividing $(11)$ by $2$, we get
$$
\int_0^\pi\log(1+\cos(x))\,\mathrm{d}x=-\pi\log(2)\tag{12}
$$
A: $$\begin{eqnarray*}\int_{0}^{\pi}\log(1+\cos x)\,dx &=& \int_{0}^{\pi/2}\log(1+\cos x)\,dx+\int_{0}^{\pi/2}\log(1+\cos(\pi-x))\,dx\\ &=& \int_{0}^{\pi/2}\log(\sin^2 x)\,dx=\int_{0}^{\pi}\log(\sin x)\,dx \tag{1}\end{eqnarray*}$$
And by a notorious identity:
$$ \prod_{k=1}^{n-1}\sin\frac{k\pi}{n} = \frac{2n}{2^n},\tag{2}$$
hence the RHS of $(1)$ can be computed as a Riemann sum:
$$ \int_{0}^{\pi}\log(\sin x)\,dx = \lim_{n\to +\infty}\frac{\pi}{n}\sum_{k=1}^{n-1}\log\sin\frac{\pi k}{n}=\color{red}{-\pi \log 2}.\tag{3}$$
There is also a well-known proof through symmetry:
$$ \begin{eqnarray*}I=\int_{0}^{\pi}\log(\sin x)&=&2\int_{0}^{\pi/2}\log(\sin(2t))\,dt=2\int_{0}^{\pi/2}\log(2\sin t\cos t)\,dt\\&=&\pi\log 2+2\int_{0}^{\pi/2}\log(\sin t)\,dt+2\int_{0}^{\pi/2}\log(\cos t)\,dt\\&=&\pi \log 2 + 2I\tag{4}\end{eqnarray*}$$
from which $I=-\pi\log 2$ immediately follows.
A: Notice, $$\color{blue}{\int_{0}^\pi\log(1+\cos x)\ dx}=2\int_{0}^{\pi/2}\log(1+\cos 2x)\ dx=2\int_{0}^{\pi/2}\log\left(2\cos^2 x\right)\ dx$$$$=4\int_{0}^{\pi/2}\log\left(\cos x\right)\ dx+2\log2\int_0^{\pi/2}\ dx$$ $$=4\color{red}{I}+\pi\log2\tag 1$$
Where, $$\color{red}{I}=\int_{0}^{\pi/2}\log\left(\cos x\right)\ dx\tag 2$$
$$I=\int_{0}^{\pi/2}\log\left(\sin \left(\frac \pi2-x\right)\right)\ dx=\int_{0}^{\pi/2}\log\left(\sin x\right)\ dx\tag 3$$
Adding (2) & (3), one should get 
$$2I=\int_{0}^{\pi/2}\log\left(\sin x\cos x\right)\ dx=\int_{0}^{\pi/2}\log\left(\frac{\sin 2x}{2}\right)\ dx=\int_{0}^{\pi/2}\log\left(\sin 2x\right)\ dx-\frac{\pi}{2}\log 2$$
$$2I=\frac{1}{2}\int_{0}^{\pi}\log\left(\sin x\right)\ dx-\frac{\pi}{2}\log 2=\frac{2}{2}\int_{0}^{\pi/2}\log\left(\sin x\right)\ dx-\frac{\pi}{2}\log 2=I-\frac{\pi}{2}\log 2$$
$$\color{red}{I=-\frac{\pi}{2}\log 2}$$, now setting the value of $I$ in (1), one should get 
$$\color{blue}{\int_{0}^\pi\log(1+\cos x)\ dx}=4\left(-\frac{\pi}{2}\log 2\right)+\pi\log 2=\color{blue}{-\pi\log 2}$$
