Integral Integral 56

From the formula for $\sinh^{-1}$ is it clear that $\sinh^{-1}(\tan x) \,=\, \ln(\sec x + \tan x)$ . Thus the substitution $u = \sec x + \tan x$ gives $\begin{aligned} \int_0^{\frac12\pi} \sinh^{-1}(\tan x)\,dx & = \; \int_0^{\frac12\pi}\ln(\sec x + \tan x)\,dx \; = \; \int_1^\infty \ln u \times \frac{2\,du}{u^2 + 1} \\ & = \; 2\int_1^\infty \frac{\ln u}{u^2 + 1}\,du \; = \; -2\int_0^1 \frac{\ln u}{u^2 + 1}\,du \; = \; 2G \end{aligned}$ making the answer $\boxed{2}$ .