Evaluate $\int_0^{{\pi}/{2}} \log(1+\cos x)\, dx$ Find the value of $\displaystyle \int_0^{{\pi}/{2}} \log(1+\cos x)\ dx$
I tried to put $1+ \cos x = 2 \cos^2 \frac{x}{2} $, but I am unable to proceed further.
I think the following integral can be helpful:
 $\displaystyle \int_0^{{\pi}/{2}} \log(\cos x)\ dx =-\frac{\pi}{2} \log2 $.

\begin{align}
\color{red}{\int^{\pi/2}_0\log(1+\cos{x}){\rm d}x}
&=\int^{\pi/2}_0\frac{x\sin{x}}{1+\cos{x}}{\rm d}x\tag1\\
&=\int^{\pi/2}_0 x\tan{\frac{x}{2}} \ {\rm d}x\tag2\\
&=4\int^{\pi/4}_0x\tan{x} \ {\rm d}x\tag3\\
&=8\sum^\infty_{n=1}(-1)^{n-1}\int^{\pi/4}_0x\sin(2nx) \ {\rm d}x\tag4\\
&=2\sum^\infty_{n=1}(-1)^{n-1}\frac{\sin\left(\frac{n\pi}{2}\right)}{n^2}-\pi\sum^\infty_{n=1}(-1)^{n-1}\frac{\cos\left(\frac{n\pi}{2}\right)}{n}\\
&=2\sum^\infty_{n=0}\frac{\sin\left(\frac{(2n+1)\pi}{2}\right)}{(2n+1)^2}-\color{grey}{2\sum^\infty_{n=1}\frac{\sin\left(n\pi\right)}{4n^2}}\\
&-\color{grey}{\pi\sum^\infty_{n=0}\frac{\cos\left(\frac{(2n+1)\pi}{2}\right)}{2n+1}}+\pi\sum^\infty_{n=1}\frac{\cos\left(n\pi\right)}{2n}\tag5\\
&=2\sum^\infty_{n=0}\frac{(-1)^n}{(2n+1)^2}+\frac{\pi}{2}\sum^\infty_{n=1}\frac{(-1)^n}{n}\tag6\\
&\large{\color{red}{=2G-\frac{\pi}{2}\log{2}}}\tag7\\
\end{align}
Explanation: 
$(1)$: Integrate by parts 
$(2)$: Trigonometric identities 
$(3)$: Substitute $x\mapsto \frac{x}{2}$ 
$(4)$: Use the Fourier series of $\tan{x}$ 
$(5)$: Split the sum into odd and even terms. It is obvious that the grey sums $=0$ 
$(6)$: $\sin\left(\frac{(2n+1)\pi}{2}\right)=\cos{n\pi}=(-1)^n$ for integer $n$. This can be verified easily. 
$(7)$: The first sum is Catalan's constant. The second is $-\log(1+1)$. (Taylor's series)
