How to prove that $\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}?$ Given that:
Where $\phi={\sqrt{5}+1\over 2}$

$$\int_{0}^{\pi/2}{\arctan(2\cos^2 x)\over \cos^2 x}\mathrm dx={\pi\over \sqrt{\phi}}\tag1$$

$t=2\cos^2 x\implies dt=-4\sin x\cos xdx$
$${1\over 4}\int_{0}^{1}{\arctan t\over t\sqrt{t-t^2}}\mathrm dt\tag2$$
$t=\tan v\implies dt=\sec^2 v dv$
$${1\over 4}\int_{0}^{\pi/4}{v\sec^2 v\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag3$$
$${1\over 2}\int_{0}^{\pi/4}{v\over \sin(2v)\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag4$$
Or we leave  it in terms of $\tan t$
$${1\over 4}\int_{0}^{\pi/4}{v(1+\tan^2 v)\over \tan v\sqrt{\tan v-\tan^2 v}}\mathrm dv\tag5$$
 A: In $(2)$, you should've gotten
$$I=\int_0^2\frac{\arctan(t)}{t\sqrt{2t-t^2}}~\mathrm dt$$
By letting $t=1/u$, we get
$$I=\int_{1/2}^{+\infty}\frac{\arctan(1/u)}{\sqrt{2u-1}}~\mathrm du$$
And by integration by parts,
$$I=\int_{1/2}^{+\infty}\frac{\sqrt{2u-1}}{1+u^2}~\mathrm du$$
which is the real part of
$$I_1=\int_{-\infty}^{+\infty}\frac{\sqrt{2u-1}}{1+u^2}~\mathrm du$$
And by taking a semi-circle contour and the residue theorem,
$$I=\Re(I_1)=\Re(\pi\sqrt{2i-1})=\frac\pi{\sqrt\phi}$$
A: Let
$$I(a)=\int_{0}^{\pi/2}{\arctan(a\cos^2 x)\over \cos^2 x}\mathrm dx.$$
Then $I(0)=0,I(2)=I$ and
\begin{eqnarray}
I'(a)&=&\int_{0}^{\pi/2}{1\over{1+a^2\cos^4 x}}\mathrm dx\\
&=&\int_{0}^{\pi/2}{\sec^2x\over{a^2+\sec^4 x}}\sec^2x\mathrm dx\\
&=&\int_{0}^{\infty}{1+u^2\over{a^2+(1+u^2)^2}}\mathrm du\\
&=&\int_{0}^{\infty}{1+u^2\over{(u^2+1+ai)(u^2+1-ai)}}\mathrm du\\
&=&\frac12\int_{0}^{\infty}\left({1\over{u^2+1+ai}}+{1\over{u^2+1-ai}}\right)\mathrm du\\
&=&\frac12\bigg[\frac{1}{\sqrt{1+ai}}\arctan\frac{u}{\sqrt{1+ai}}+\frac{1}{\sqrt{1-ai}}\arctan\frac{u}{\sqrt{1-ai}}\bigg]\bigg|_{0}^{\infty}\\
&=&\frac{\pi}{4}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)
\end{eqnarray}
So
\begin{eqnarray}
I(2)&=&\frac{\pi}{4}\int_{0}^{2}\bigg(\frac{1}{\sqrt{1-ai}}+\frac{1}{\sqrt{1+ai}}\bigg)\\
&=&\frac{\pi}{4}\bigg(\frac{\sqrt{1+ai}}{i}+\frac{\sqrt{1-ai}}{-i}\bigg)\bigg|_0^2\\
&=&\frac{\pi}{4}\frac{\sqrt{1+2i}-\sqrt{1-2i}}{i}\\
&=&\frac{\sqrt{2\sqrt5-2}}{4}\pi=\frac{\pi}{\sqrt\phi}
\end{eqnarray}
