Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$

Question

Discovered the integral $$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$ which looks clean, yet challenging. Have not seen it before; and shared here in case of interest.

Edit:

Let $J(a)=\int_0^{\frac\pi{3}}\tanh^{-1}\frac{2a\sin x}{1+a^2}dx$ $$ J’(a) = \int_0^{\frac\pi{3}}\frac{2(1-a^2)\sin x}{4a^2\cos^2x+(1-a^2)^2}dx =\frac1a\tan^{-1}\frac {a(1-a^2)}{1+a^4} $$ Then \begin{align} I & =J(1) =\int_0^1 J’(a)da = \int_0^1\frac1a \tan^{-1}\frac {a(1-a^2)}{1+a^4}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}

Answer 1

$$\int_0^\frac{\pi}{3} \operatorname{arctanh}(\sin x)dx\overset{\sin x\to x}=\int_0^\frac{\sqrt 3}{2} \frac{\ln \left(\frac{1+x}{1-x}\right)}{\sqrt{1-x^2}}dx\overset{\large \frac{1-x}{1+x}\to x^2}=-2\int_{2-\sqrt 3}^1\frac{\ln x}{1+x^2}dx$$ $$=-2\underbrace{\int_0^1\frac{\ln x}{1+x^2}dx}_{-G}+2\underbrace{\int_0^{2-\sqrt 3}\frac{\ln x}{1+x^2}dx}_{x\to\tan x}=2G+2\int_0^\frac{\pi}{12}\ln(\tan x)dx=\frac23G$$ See here for the last integral.

Answer 2

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

Answer 3

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!