Integral Integral 57

Note that $\tanh^{-1}(\cos x) \; = \; \tfrac12\ln\left(\frac{1 + \cos x}{1 - \cos x}\right) \; = \; \tfrac12\ln\big(\cot^2\tfrac12x\big) \; = \; \ln\big(\cot\tfrac12x\big)$ and hence $\begin{aligned} \int_0^{\frac12\pi}\tanh^{-1}(\sin x)\,dx & = \; \int_0^{\frac12\pi}\tanh^{-1}(\cos x)\,dx \; = \; \int_0^{\frac12\pi} \ln\big(\cot\tfrac12x\big)\,dx \; = \; -2\int_0^{\frac14\pi}\ln(\tan x)\,dx \\ & = \; -2\int_0^1 \frac{\ln u}{1 + u^2}\,du \; = \; 2\sum_{n=0}^\infty (-1)^{n-1}\int_0^1 u^{2n} \ln u\,du \; = \; 2\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} \; = \; 2G \end{aligned}$ making the answer $\boxed{2}$ .