Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tanh^{-1}\frac{2a\sin x}{1+a^2}dx$, with $I(0)=0$
$$ J’(a) = \int_0^{\frac\pi{3}}\frac{2(1-a^2)\sin x}{4a^2\cos^2x+(1-a^2)^2}dx
=\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a}
$$
Then
\begin{align}
I
& =J(1) =J(0)+\int_0^1 J’(a)da = \int_0^1\frac{\tan^{-1}\frac {a(1-a^2)}{1+a^4}}{a} da\\
&=\int_0^1\left(\frac{\tan^{-1}a}{a}\right.
-\underset{a^3\to a}{\left.\frac{\tan^{-1}a^3}{a}\right)}da=\left(1-\frac13\right) \int_0^1\frac{\tan^{-1}a}{a}da=\frac23G
\end{align}
 A: Note that $2\operatorname{arctanh} x=\ln\left(\frac{1+x}{1-x}\right)$. So by letting $\frac{1-\sin x}{1+\sin x}=y\,$ in the integral we get: $$\int_0^\frac{\pi}{3} \operatorname{arctanh}(\sin x)dx=-\int_0^{(2-\sqrt 3)^2} \frac{\ln y}{\sqrt y(1+y)}dy\overset{\sqrt y=\tan x}=-\int_0^\frac{\pi}{12}\ln(\tan x)=\frac23G$$
See here for the last integral.
A: Use that 
$$\operatorname{arctanh}(\sin(x))=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{\sin((2n-1)x)}{2n-1}, \ 0 <x<2\pi$$and after integration, employ the result in $(3.238)$, page $215$, from the book (Almost) Impossible Integrals, Sums, and Series$\displaystyle \left(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(\pi/6(4n+1))}{(2n-1)^2}=\sum_{n=1}^{\infty} \frac{\sin(\pi/6(2n-1))}{(2n-1)^2}=\frac{2}{3}G\right)$.
A: Let $I$ be the integral given by
$$\begin{align}
I&=\int_0^{\pi/3}\text{arctanh}(\sin(x))\,dx\\\\
&=\frac12\int_0^{\pi/3}\log\left(\frac{1+\sin(x)}{1-\sin(x)}\right)\,dx\\\\
&=\frac12\int_{-\pi/3}^{\pi/3}\log\left(1+\sin(x))\right)\,dx\tag1
\end{align}$$
Then, enforcing the substitution $x\mapsto \pi/2-x$ in $(1)$, we find that
$$\begin{align}
I&=\frac12\int_{\pi/6}^{\pi/2}\log\left(\frac{1+\cos(x)}{1-\cos(x)}\right)\,dx\\\\
&=-\int_{\pi/6}^{\pi/2}\log\left(\tan(x/2)\right)\,dx\\\\
&=-2\int_{\pi/12}^{\pi/4}\log\left(\tan(x)\right)\,dx\tag2
\end{align}$$
Next, using the Fourier series for $\tan(x)=\sum_{k=1}^\infty \frac{(-1)^k-1}{k}\cos(2kx)$ in $(2)$ reveals
$$\begin{align}
I&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)-\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)-\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{2}\right)}{(2k-1)^2}-2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\sum_{k=1}^\infty \frac{(-1)^k}{(2k-1)^2}-2\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\\\\
&=2\left(G-\sum_{k=1}^\infty \frac{\sin\left(\frac{(2k-1)\pi}{6}\right)}{(2k-1)^2}\right)\tag3
\end{align}$$
Using judicious grouping, user @M.N.C.E. showed in THIS ANSWER that the right-hand side of $(3)$ was equal to $2G/3$.  Hence  we find that 
$$I=2G/3$$
as was to be shown!
