Question: How can I show that \begin{align} & \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \text{arctanh}\left(\sin(2x)\right)\,dx \\[2mm] = & \ {\small\log\left(\left(2\sqrt{2-\sqrt{2}}+2\sqrt{2}-1\right)^{\sqrt{2+\sqrt{2}}} \left(1+2\sqrt{2}-2\sqrt{2+\sqrt{2}}\right)^{\sqrt{2-\sqrt{2}}}\right)} \end{align}
My attempt
I will rewrite $\text{arctanh}(\sin(2x))$ in terms of logarithms.
We know that $$ \text{arctanh}(y) = \frac{1}{2} \ln\left(\frac{1+y}{1-y}\right) $$
Here $y=\sin(2x)$ $$ \text{arctanh}(\sin(2x)) = \frac{1}{2} \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) $$
Using trigonometric identities, we can write: $$ 1 + \sin(2x) = 1 + 2 \sin(x) \cos(x) $$ $$ 1 - \sin(2x) = 1 - 2 \sin(x) \cos(x) $$
or
$\sin(2x) = 2 \sin(x) \cos(x)$
Substitute $\text{arctanh}(\sin(2x))$: $$ \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \text{arctanh}(\sin(2x)) \, dx \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \cdot \frac{1}{2} \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) \ dx $$
Simplifying the logarithm expression:
Using the given trigonometric identities: $$ 1 + \sin(2x) = \left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2 $$ $$ 1 - \sin(2x) = \left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2 $$
Thus, $$ \ln\left(\frac{1+\sin(2x)}{1-\sin(2x)}\right) = \ln\left(\frac{\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2}{\left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2}\right) $$
so our integral becomes
$$ \int_0^{\frac{\pi}{2}} \sin\left(\frac{x}{2}\right) \ln\left(\frac{\left(\sin\left(\frac{x}{2}\right) + \cos\left(\frac{x}{2}\right)\right)^2}{\left(\sin\left(\frac{x}{2}\right) - \cos\left(\frac{x}{2}\right)\right)^2}\right)\, dx $$