TITLE: Evaluate $\int_0^{{\pi}/{2}} \log(1+\cos x)\, dx$
QUESTION [7 upvotes]: 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 $.

REPLY [2 votes]: Let's try your way. Put $1+\cos x=2\cos^2\frac x2$ and use substitution $\frac x2\mapsto x$, we get
\begin{equation}
\int_0^{{\pi}/{2}} \log(1+\cos x)\ dx=\int_0^{{\pi}/{2}} \log2\ dx +4\int_0^{{\pi}/{4}} \log \cos x\ dx
\end{equation}
The first integral is a piece of cake and the second integral has been evaluated here. The final result is
\begin{equation}
\int_0^{{\pi}/{2}} \log(1+\cos x)\ dx=\frac{\pi}{2} \log2 +4\left(\frac C2-\frac{\pi}{4} \log2\right)=2C-\frac{\pi}{2} \log2
\end{equation}