Area

We need the two integrals $\begin{array}{rcl} I_1 & = & \displaystyle \int_0^{\frac12\pi} x \sqrt{\tan x}\,dx \; = \; \int_0^{\frac12\pi} \frac{x \sin x\,dx}{\sqrt{\sin x \cos x}} \; = \; \sqrt{2}\int_0^{\frac12\pi} \frac{x \sin x}{\sqrt{\sin 2x}}\,dx \\ & = & \displaystyle \frac{1}{2\sqrt{2}}\int_0^\pi \frac{x \sin \tfrac12x}{\sqrt{\sin x}}\,dx \; = \; -\frac{1}{2\sqrt{2}}\frac{\partial F}{\partial a}\big(\tfrac12,\tfrac12\big) \; = \; -\frac{1}{2\sqrt{2}} G'(\tfrac12)\;, \\ I_2 & = & \displaystyle \int_0^{\frac12\pi} x^2 \sqrt{\cot x}\,dx \; = \; \int_0^{\frac12\pi} \frac{x^2 \cos x\,dx}{\sqrt{\sin x \cos x}} \; = \; \sqrt{2} \int_0^{\frac12\pi} \frac{x^2 \cos x}{\sqrt{\sin2x}}\,dx \\ & = & \displaystyle \frac{1}{4\sqrt{2}}\int_0^\pi \frac{x^2 \cos \tfrac12x}{\sqrt{\sin x}}\,dx \; = \; -\frac{1}{4\sqrt{2}} \frac{\partial^2 F}{\partial a^2}\big(\tfrac12,\tfrac12\big) \; = \; -\frac{1}{4\sqrt{2}} G''(\tfrac12)\;, \end{array}$ where we are using the standard integral $F(\nu,a) \; = \; \int_0^\pi \sin^{\nu-1}x\, \cos ax\,dx \; = \; \frac{\pi \cos\big(\frac12 a \pi\big)}{2^{\nu-1} \nu B\big(\frac12(\nu+a+1),\frac12(\nu-a+1)\big)} \;,$ valid for $\nu > 0$ , and so $G(a) \; = \; F(\tfrac12,a) \; =\; \frac{2^{\frac32} \pi \cos(\tfrac12 a \pi)}{B\big(\tfrac{3+2a}{4},\tfrac{3-2a}{4}\big)} \;.$ Both integrals can be evaluated in terms of polygamma functions, and then simplified, yielding $\begin{array}{rcl} \displaystyle I_1 \; = \; \int_0^{\frac12\pi} x\sqrt{\tan x}\,dx & = & \displaystyle \frac{\pi}{4\sqrt{2}}\big[\pi + \ln4\big] \\ \displaystyle I_2 \; = \; \int_0^{\frac12\pi} x^2\sqrt{\cot x}\,dx & = & \displaystyle \frac{\pi}{48\sqrt{2}}\big[5 \pi^2 - 6 \pi \ln4 - 3 (\ln4)^2\big]\;. \end{array}$ and hence $\int_0^{\frac12\pi} f(x)\,dx \; =\; I_1 + I_2 \; =\; \frac{\pi}{4\sqrt{2}}\big[\pi + \ln4\big] \,+\, \frac{\pi}{48\sqrt{2}}\big[5 \pi^2 - 6 \pi \ln4 - 3 (\ln4)^2\big] \; = \; \boxed{3.32227}$