$$\int_{0}^{\pi/2}\arctan\left({2\over \cos^2{x}}\right)\mathrm dx=\pi\arctan\left({1\over \sqrt{\phi}}\right)\tag1$$
$\phi$ is the golden ratio
$2\sec^2{x}=2\tan^2{x}+2$
$u=\sec^2{x}$ then $\mathrm du=2{\tan{x}\over \cos^2{x}}\mathrm dx$
$$\int{\cos^2{x}\over \tan{x}}\arctan\left({2\over u}\right)\mathrm du\tag2$$
$$\int{1\over u\sqrt{u-1}}\arctan\left({2\over u}\right)\mathrm du\tag3$$
Not so sure what is next step...
How do we show that $(1)=\pi\arctan\left({1\over \sqrt{\phi}}\right)$
$$\begin{eqnarray*}\int_{0}^{\pi/2}\arctan\left(\frac{2}{\cos^2\theta}\right)\,d\theta&=&\int_{0}^{\pi/2}\arctan\left(\frac{2}{\sin^2\theta}\right)\,d\theta\\&=&\int_{0}^{1}\frac{\arctan\frac{2}{u^2}}{\sqrt{1-u^2}}\,du\\&=&\int_{0}^{1}\frac{\frac{\pi}{2}-\arctan\frac{u^2}{2}}{\sqrt{1-u^2}}\,du\\&=&\frac{\pi^2}{4}-\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,2^{2n+1}}\int_{0}^{1}\frac{u^{4n+2}}{\sqrt{1-u^2}}\,du\\&=&\frac{\pi^2}{4}-\sum_{n\geq 0}\frac{\pi(-1)^n (4n+1)}{(2n+1)^22^{6n+3}}\binom{4n}{2n}\end{eqnarray*}\tag{A}$$
The last hypergeometric series can be computed from
$$ \sum_{n\geq 0}\frac{2n+1}{(n+1)^2}\binom{2n}{n}z^n =\frac{\log 2-\log\left(1+\sqrt{1-4z}\right)}{z}\tag{B} $$
leading to
$$ \sum_{n\geq 0}\frac{4n+1}{(2n+1)^2}\binom{4n}{2n}z^n =\frac{1}{2\sqrt{z}}\,\log\left(\frac{1+\sqrt{1+4\sqrt{z}}}{1+\sqrt{1-4\sqrt{z}}}\right) \tag{C}$$
$$ \sum_{n\geq 0}\frac{z^n\,\Gamma\left(2n+\tfrac{3}{2}\right)}{(2n+1)\,\Gamma(2n+2)}=\sqrt{\frac{\pi}{z}}\,\text{arctanh}\sqrt{\frac{1-\sqrt{1-z}}{2}}\tag{D}$$
and finally to
$$ \int_{0}^{\pi/2}\arctan\left(\frac{2}{\cos^2\theta}\right)\,d\theta = \pi\arctan\sqrt{\frac{2}{1+\sqrt{5}}}.$$
As a corollary, $z=\frac{1-\sqrt{5}}{2}$ is a special value for the hypergeometric function ${}_2 F_1\left(\tfrac{3}{4},\tfrac{3}{4};\tfrac{7}{4};z\right)$.
This can be done also by differentiation under the integral sign, since $\int_{0}^{\pi/2}\frac{\partial}{\partial a}\arctan\left(\frac{a}{\cos^2\theta}\right)\,d\theta$ is not that difficult to compute.
