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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.

enter image description here

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 $I=\frac{\pi}{2}\ln{2}$.

Dan
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    Note that $\int_{0}^{\frac{\pi}{2}}\ln\left(\sqrt{\cos\left(x\right)+1}+\sqrt{\cos\left(x\right)}\right)dx = \int_{0}^{\frac{\pi}{2}}\sinh^{-1}\left(\sqrt{\cos\left(x\right)}\right)dx = \frac{1}{2}\int_{0}^{\pi}\sinh^{-1}\left(\sqrt{\sin\left(x\right)}\right)dx$, and the answer for the last integral is found here. I'll see if I can come up with my own solution, but right now my attempt is a mess. – Accelerator Oct 23 '22 at 10:08
  • @Accelerator Did you find easy way? – Bob Dobbs Oct 24 '22 at 07:33
  • @BobDobbs No. This was all I was able to do: https://www.desmos.com/calculator/exa5bmiqo3 – Accelerator Oct 24 '22 at 07:36

1 Answers1

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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 $\int_0^{\pi/2}\ln(\tan x +1-\sqrt{2\tan x})dx=0$

Quanto
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