Evaluate $\int_{-\pi/4}^{\pi/4}\frac{x}{\sin x}\mathrm{d}x$ I am working on the integral 
$$I=\int_{-\pi/4}^{\pi/4}\frac{x}{\sin x}\mathrm{d}x=2\int_0^{\pi/4}\frac{x}{\sin x}\mathrm{d}x$$
Which I am fairly confident has a closed form, as $$\int_{0}^{\pi/2}\frac{x}{\sin x}\mathrm{d}x=2G$$
Where $G$ is Catalan's constant.
Preforming a tangent half angle substitution, we have that
$$I=4\int_0^{\sqrt{2}-1}\frac{\arctan x}{x}\mathrm{d}x$$
Then using $$\arctan x=\sum_{n\geq0}(-1)^n\frac{x^{2n+1}}{2n+1}$$
We have 
$$I=4\sum_{n\geq0}\frac{(-1)^n}{(2n+1)^2}(\sqrt{2}-1)^{2n+1}$$
Which is painfully similar to $G$. I do not know how to deal with that extra $(\sqrt{2}-1)^{2n+1}$ bit though...
In another post of mine I showed that 
$$I=\pi\sum_{n\geq1} n\log\bigg(\frac{4n+1}{4n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}$$
And similarly I showed that 
$$\sum_{n\geq1}n\log\bigg(\frac{2n+1}{2n-1}\bigg)\prod_{k\geq1\\k\neq n}\frac{k^2}{k^2-n^2}=\frac{4G}\pi$$
So I have to questions. How do I find an exact value for $I$? And are the last two series representations correct? Thanks.
Major Edit:
Okay so I found a closed form for the integral. Wolfy gave me
$$\int\frac{x}{\sin x}\mathrm{d}x=i\bigg(\text{Li}_2(-e^{ix})-\text{Li}_2(e^{ix})\bigg)+x\log\frac{1-e^{ix}}{1+e^{ix}}$$
I guess that Wolfy didn't want to do the algebra, so I did it by hand. It took me like $10$ minutes, but I am pretty sure that
$$I=-\frac34G+\frac{\pi^2}4\bigg(\frac{13}{24}-i\bigg)-\frac{i\pi}4\log(1+\sqrt{2})+\frac{i-1}{32\sqrt{2}}\bigg[\psi^{(1)}\bigg(\frac{5}{8}\bigg)-\psi^{(1)}\bigg(\frac{1}{8}\bigg)\bigg]+\frac{i+1}{32\sqrt{2}}\bigg[\psi^{(1)}\bigg(\frac{3}{8}\bigg)-\psi^{(1)}\bigg(\frac{7}{8}\bigg)\bigg]$$
Where $\psi^{(1)}$ is the first derivative of the di-gamma function.
 A: For the purpose of an alternative method,  mostly relying on Clausen function. We have:
$$I=\int_{-\pi/4}^{\pi/4}\frac{x}{\sin x} dx=2\int_0^{\pi/4}x\left(\ln\left(\tan\frac{x}{2}\right)\right)' dx=2x\ln\left(\tan\frac{x}{2}\right)\bigg|_0^\frac{\pi}{4}-2\int_0^\frac{\pi}{4}\ln\left(\tan\frac{x}{2}\right)dx=$$
$$=\frac{\pi}{2}\ln(\sqrt 2-1)-2\int_0^\frac{\pi}{4}\left(\ln\left(2\sin\frac{x}{2}\right)-\ln\left(2\cos\frac{x}{2}\right)\right)dx=$$$$=\frac{\pi}{2}\ln(\sqrt 2-1)+2\text{Cl}_2\left(\frac{\pi}{4}\right)+2\text{Cl}_2\left(\frac{3\pi}{4}\right)$$
The last two integrals can be found on the first link. Now is up to the reader if using Clausen function gives any satisfaction, because in desguise is still a series, but same goes with the trigamma function.
An interesting question might be: For what values of $\phi$  does the following integral have an elementary answer (Catalan's constant included)?
$\ \displaystyle{I(\phi)=\int_0^\phi \frac{x}{\sin x}dx}$. So far I only know about $I\left(\frac{\pi}{6}\right)$.
A: Too long for comments.
$$I=4\sum_{n\geq0}\frac{(-1)^n}{(2n+1)^2}(\sqrt{2}-1)^{2n+1}=4 \left(\sqrt{2}-1\right) \,
   _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},\frac{3}{2};2
   \sqrt{2}-3\right)$$
On the other hand, working from Wolfram Alpha expression for the antiderivative and trying to simplify as much as I could  the integral, I got
$$I=-\frac{\pi}{4} \,  \log \left(3+2 \sqrt{2}\right)+\frac 1 {16 \sqrt 2} \left(\psi ^{(1)}\left(\frac{1}{8}\right)+\psi ^{(1)}\left(\frac{3}{8}\right)-\psi
   ^{(1)}\left(\frac{5}{8}\right)-\psi ^{(1)}\left(\frac{7}{8}\right) \right)$$
What is interesting to mention is that, doing calculations with another CAS, intermediate steps show $G$ appearing a few times.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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\begin{align}
I & \equiv \int_{-\pi/4}^{\pi/4}{x \over \sin\pars{x}}\,\dd x,\
\pars{~\mbox{OP already shows that}\
I = \bbox[10px,#ffd]{\!\!\!\!\! 4\!\int_{0}^{\root{2} - 1}\!\!{\arctan\pars{x} \over x}\,\dd x}~}
\end{align}

Then,
\begin{align}
I & = \bbox[10px,#ffd]{4\int_{0}^{\root{2} - 1}{\arctan\pars{x} \over x}\,\dd x} =
4\,\Im\int_{0}^{\root{2} - 1}{\ln\pars{1 + \ic x} \over x}\,\dd x
\\[5mm] &
\stackrel{{\large x\ =\ \ic t} \atop {\large t\ =\ -\ic x}}{=}\,\,\,
4\,\Im\int_{0}^{-\pars{\root{2} - 1}\ic}
{\ln\pars{1 - t} \over  t}\,\dd t
\\[5mm] & =
-4\,\Im\int_{0}^{-\pars{\root{2} - 1}\ic}
\mrm{Li}_{2}'\pars{t}\,\dd t =
-4\,\Im\mrm{Li}_{2}\pars{-\bracks{\root{2} - 1}\ic}
\\[5mm] & =
\bbx{4\,\Im\mrm{Li}_{2}\pars{\bracks{\root{2} - 1}\ic}}
\approx 1.6271
\end{align}
A: \begin{align}I&=2\int_0^{\pi/4}\frac{x}{\sin x}\mathrm{d}x\\
&=2\Big[x\ln\left(\tan\left(\frac{x}{2}\right)\right)\Big]_0^{\pi/4}-2\int_0^{\pi/4}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\
&=\frac{\pi}{2}\ln(\sqrt{2}-1)-2\int_0^{\pi/4}\ln\left(\tan\left(\frac{x}{2}\right)\right)\,dx\\
\end{align}
In the latter integral perform the change of variable $y=\dfrac{x}{2}$,
\begin{align}I&=\frac{\pi}{2}\ln(\sqrt{2}-1)-4\int_0^{\pi/8}\ln\left(\tan\left(x\right)\right)\,dx\\
&=\frac{\pi}{2}\ln(\sqrt{2}-1)-4\Big[\frac{1}{2}\text{i}\big(\text{Li}_2(\text{i}\tan x)-\text{Li}_2(-\text{i}\tan x)\big)+x\ln\left(\tan x\right)\Big]_0^{\pi/8}\\
&=\boxed{2\text{i}\left(\text{Li}_2\left(\text{i}\left(1-\sqrt{2}\right)\right)-\text{Li}_2\left(\text{i}\left(\sqrt{2}-1\right)\right)\right)}
\end{align}
