calculus - 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 $.

Answer

Using Weierstrass substitution
$$
t=\tan\frac x2\qquad;\qquad\cos x=\frac{1-t^2}{1+t^2}\qquad;\qquad dx=\frac{2}{1+t^2}\ dt
$$
we obtain
\begin{align}
\int_0^{\Large\frac\pi2}\ln(1+\cos x)\ dx&=2\underbrace{\int_0^1\frac{\ln2}{1+t^2}\ dt}_{\color{blue}{\text{set}\ t=\tan\theta}}-2\color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}\\
&=\frac{\pi}{2}\ln2-2\color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}.\tag1
\end{align}
Consider

\begin{align}
\int_0^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt&=\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt+\underbrace{\int_1^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}_{\large\color{blue}{t\ \mapsto\ \frac1t}}\\
&=2\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt-2\int_0^1\frac{\ln t}{1+t^2}\ dt\\
\color{red}{\int_0^1\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}&=\frac12\underbrace{\int_0^\infty\frac{\ln\left(1+t^2\right)}{1+t^2}\ dt}_{\color{blue}{\text{set}\ t=\tan\theta}}+\int_0^1\frac{\ln t}{1+t^2}\ dt\\
&=-\underbrace{\int_0^{\Large\frac\pi2}\ln\cos\theta\ d\theta}_{\color{blue}{\Large\text{*}}}+\sum_{k=0}^\infty(-1)^k\underbrace{\int_0^1 t^{2k}\ln t\ dt}_{\color{blue}{\Large\text{**}}}\\
&=\frac\pi2\ln2-\sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)^2}\\
&=\frac\pi2\ln2-\text{G},\tag2
\end{align}
where $\text{G}$ is Catalan's constant.

$(*)$ can be proven by using the symmetry of $\ln\cos\theta$ and $\ln\sin\theta$ in the interval $\left[0,\frac\pi2\right]$ and $(**)$ can be proven by using formula
$$
\int_0^1 x^\alpha \ln^n x\ dx=\frac{(-1)^n n!}{(\alpha+1)^{n+1}}, \qquad\text{for }\ n=0,1,2,\ldots
$$
Thus, plugging in $(2)$ to $(1)$ yields
\begin{align}
\int_0^{\Large\frac\pi2}\ln(1+\cos x)\ dx
=\large\color{blue}{2\text{G}-\frac{\pi}{2}\ln2}.
\end{align}