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}

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$\displaystyle \left(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(\pi/6(4n+1))}{(2n-1)^2}=\sum_{n=1}^{\infty} \frac{\sin(\pi/6(2n-1))}{(2n-1)^2}=\frac{2}{3}G\right)$.
