Aww!! That's The Beauty

The quadratic equation $z^2 - 2(1+2i)z + 1 = 0$ has zeros $\lambda_\pm = 1 + 2i \pm 2^{\frac54}e^{\frac{3\pi i}{8}}$ , where $|\lambda_+| > 1 > |\lambda_-|$ , and hence $\int_0^{2\pi} \frac{d\theta}{\cos\theta - 1 - 2i} \; = \; \int_{|z|=1} \frac{2}{z+z^{-1}-2(1+2i)}\,\frac{dz}{iz} \; = \; \frac{2}{i}\int_{|z|=1}\frac{dz}{z^2 - 2(1+2i)z+1} \; = \; \frac{4\pi}{\lambda_- - \lambda_+} \; = \; -\frac{\pi}{2^{\frac14}}e^{-\frac{3\pi i}{8}}$ by standard Residue Calculus tricks. Thus, with the substitutions $u= \tan x$ , $u = \sin^2\theta$ , followed by a move to double angles, partial fractions and finally the above result, we obtain that $\begin{aligned} \int_0^{\frac14\pi} \sqrt{\tan x (1 - \tan x)}\,dx & = \; \int_0^1 \frac{\sqrt{u(1-u)}}{1+u^2}\,du \; = \; \int_0^{\frac12\pi} \frac{2\sin^2\theta \cos^2\theta}{1 + \sin^4\theta}\,d\theta \; = \; 2\int_0^{\frac12\pi}\left(\frac{1 + \sin^2\theta}{1 + \sin^4\theta} - 1\right)\,d\theta \\ & = \; 4\int_0^{\frac12\pi}\frac{3 - \cos2\theta}{4 + (1 - \cos2\theta)^2}\,d\theta - \pi \; = \; 2\int_0^\pi \frac{3 - \cos\theta}{4 + (1 - \cos\theta)^2}\,d\theta - \pi \\ & = \; \int_0^{2\pi} \frac{3 - \cos\theta}{4 + (1 - \cos\theta)^2}\,d\theta - \pi \; = \; -\tfrac12(1+i)\int_0^{2\pi} \frac{d\theta}{\cos\theta - 1 - 2i} - \tfrac12(1-i)\int_0^{2\pi} \frac{d\theta}{\cos\theta - 1 + 2i} - \pi \\ & = \; -\mathfrak{Re}\left[(1 + i)\int_0^{2\pi}\frac{d\theta}{\cos\theta - 1 - 2i}\right] - \pi \; = \; \mathfrak{Re}\left[ \sqrt{2}e^{\frac{\pi i}{4}} \frac{\pi}{2^{\frac14}}e^{-\frac{3\pi i}{8}}\right] - \pi \; = \; 2^{\frac14}\pi \cos \tfrac18\pi - \pi \\ & = \; \pi\left(\sqrt{\tfrac12 + \tfrac{1}{\sqrt{2}}} - 1\right) \end{aligned}$ which makes $\phi_1 + \phi_2 + \phi_3 = 60 + 45 + 0 = \boxed{105}$ .