Long story short. I was investigating Coxeter's\Ahmed's integrals; I scoured through the Internet, documenting every integral related to the subject, and currently I hit a road block, not knowing how to proceed.
First, I should discuss shorty what I have done. Following the method of Pisco in this post (Which comes from a paper published by Coxeter found in this book.), putting $$ \mathcal{A} \left( a,b,c \right) =\int_0^c{\sqrt{\frac{a-b}{b-x}}\frac{\mathrm{arctan} \left( \sqrt{x} \right)}{a-x}\mathrm{d}x} $$ where $a\geqslant b\geqslant c\geqslant 0$.
$\mathcal{A}$ can be studied by turning it into a function $\mathcal{S}$ introduced by Schläfli, as Pisco demostrated. The detail is of this is collected and sorted by me here. With this tool, I was able to evaluate almost every related question I found online, except two, and they are the roadblock I encountered.
Question 1:
The following integral is first proposed by Coxeter in 1988 American Mathematics Monthly.
$$
\int_1^6{\frac{\mathrm{arcsec} \left( x \right)}{\left( x+2 \right) \sqrt{x+1}}\left( \frac{1}{\sqrt{x+3}}+2 \right) \mathrm{d}x}
$$
The value of which is $\frac{2\pi^2}{15}$. It has been asked here with no answer. What I want to ask here is NOT how to evaluate this integral, but rather, its relation with another integral.
In this post by Sangchul Lee (as well as in his articles regerding to generalising Ahmed's integral), he mentioned the integral atop is directly connected to
$$
\int_0^{\frac{\pi}{3}}{\mathrm{arccos} \left( \frac{\cos \left( x \right)}{1+2\cos \left( x \right)} \right)}\mathrm{d}x
$$
This happens to be one of the integrals I have calculated, the value of which is also $\frac{2\pi^2}{15}$. In spite of that, I cannot see said connection other than they have the same value. On top of that, in the same post, I also cannot see the connection between the last integral and the equation
$$\int_0^{\pi/3} \arctan (\sqrt{5} \tan \theta) \, d\theta - 2 \int_0^{\pi/6} \arctan (\sqrt{5} \tan \theta) \, d\theta = \frac{\pi^{2}}{30}$$
Please illuminate me what the relation is between these integrals, because for me, the connection between them is difficult to fathom with standard methods.
Question 2:
This one is simpler in terms of what I want to know. Is the following integral can be evaluated by methods developed by either Coxeter or Sangchul Lee?
$$
\int_0^{\frac{\pi}{3}}{\mathrm{arccos} \left( \frac{\cos \left( x \right) -1}{4\cos \left( x \right) +1} \right) \mathrm{d}x}
$$
I found it here, question 36, and appearently it is from Sangchul Lee himself. It would suffice to write it in terms of $\mathcal{A}$, if that is easier than just evaluate it.
What I have done with regards to each question:
For the first one, I have simplified it as follows
\begin{align*}
\int_1^6{\frac{\mathrm{arcsec} \left( x \right)}{\left( x+2 \right) \sqrt{x+1}}\left( \frac{1}{\sqrt{x+3}}+2 \right) \mathrm{d}x}
=&2\pi \mathrm{arcsec} \left( 6 \right) -\int_1^6{\frac{2}{x\sqrt{x^2-1}}\left( \pi -\mathrm{arccos} \left( \frac{x}{x+2} \right) +\frac{1}{2}\mathrm{arccos} \left( \frac{1}{x+2} \right) \right) \mathrm{d}x}
\\
=&\int_0^{\mathrm{arcsec} \left( 6 \right)}{\left( 2\mathrm{arccos} \left( \frac{1}{1+2\cos \left( t \right)} \right) -\mathrm{arccos} \left( \frac{\cos \left( t \right)}{1+2\cos \left( t \right)} \right) \right) \mathrm{d}t}
\\
=&\frac{2\pi ^2}{3}-\pi \mathrm{arcsec} \left( 6 \right) -4\mathcal{A} \left( 1,\frac{1}{2},\frac{1}{7} \right) -2\mathcal{A} \left( 3,2,\frac{9}{7} \right)
\\
=&4\mathcal{A} \left( 3,1,\frac{5}{7} \right) +2\mathcal{A} \left( 3,2,\frac{5}{7} \right)
\end{align*}
The crux here is these two number are not listed as special values of $\mathcal{A}$, on top of that, they are not even in the same orbit (from a group action); and even if they are in the same orbit, the coefficient is different, meaning that the non-special value part cannot be fully cancelled out. Therefore, I am stuck.
As for the second question, I have tried the method I studied, but I cannot write it in a combination of $\mathcal{A}$ values. If this is possible, it could be a lot more easier. On a side note, I can evaluate $$ \int_0^{\frac{\pi}{3}}{\mathrm{arccos}\left( \frac{\cos \left( x \right) -\color{red}{2}}{4\cos \left( x \right) +1} \right) \mathrm{d}x} $$ and it equals to $\pi \mathrm{arctan} \left( \sqrt{15} \right) -\frac{2}{9}\pi ^2$
Any help is appriciated, thanks.
