I'm trying to evaluate $I=\int_0^{\pi/2}\ln{(\sqrt{\cos{x}+1}+\sqrt{\cos{x}})}dx$.
My attempt:
I have tried using the facts that
$I=\int_0^{\pi/2}\ln{(\sqrt{\sin{x}+1}+\sqrt{\sin{x}})}dx$, and
$I=-\int_0^{\pi/2}\ln{(\sqrt{\cos{x}+1}-\sqrt{\cos{x}})}dx$
and then adding these various forms together.
I have also tried finding a substitution that will convert $I$ into an integral of an odd function , as well as integration by parts. All without success.
Context:
This is part of my attempt to solve the following problem, which I made up.
The diagram shows a quarter-circle of radius $2$ and line segments of lengths $l_0, l_1, l_2, …, l_n$ with equal angles between them. Two of the line segments are tangent to the quarter-circle at the ends of the quarter-circle. Show that $\lim\limits_{n\to\infty}\prod\limits_{k=0}^n l_k =2$.
This amounts to showing that $I=\frac{\pi}{2}\ln{2}$.
Best Answer
Note that \begin{align} I=&\int_0^{\pi/2}\ln{(\sqrt{\sin{x}+1}+\sqrt{\sin{x}})}dx\\ =&\ -\frac12\int_0^{\pi}\ln{(\sqrt{\sin{x}+1}-\sqrt{\sin{x}})}\ dx\\ =&\ -\frac12\int_0^{\pi}\ln{\frac{1+\tan{\frac x2}-\sqrt{2\tan{\frac x2}}}{\sqrt{1+\tan^2\frac x2}}} \ \overset{\frac x2\to x} {dx}\\ =& -\int_0^{\pi/2}\ln(\cos x )dx - \int_0^{\pi/2}\ln(\tan x +1-\sqrt{2\tan x})dx\\ =& \ \frac\pi2\ln2 +0 \end{align} where $\int_0^{\pi/2}\ln(\tan x +1-\sqrt{2\tan x})dx=0$