How does one show that $\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}?$ Consider

$$\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}\tag1$$
  $\phi$;Golden ratio

An attempt:
$x=\sin{y}$ then $\mathrm dx=\cos{y}\mathrm dy$
$(1)$ becomes
$$\int_{0}^{\pi/2}{\mathrm dy\over 1+\phi \sin^4{y}}\tag2$$
Apply Geometric series to $(2)$,
$$\int_{0}^{\pi/2}(1-\phi\sin^4{y}+\phi^2\sin^8{y}-\phi^3\sin^{12}{y}+\cdots)\mathrm dy\tag3$$
$${\pi\over 2}-\int_{0}^{\pi/2}(\phi\sin^4{y}-\phi^2\sin^8{y}+\phi^3\sin^{12}{y}-\cdots)\mathrm dy\tag4$$
Power of sine seem difficult to deal with
How else can we tackle $(1)?$

A further substitution $\tan y=x$ makes the integral into,
$\displaystyle I = \dfrac{1}{\phi+1}\int\limits_0^\infty \dfrac{x^2+1}{x^4+\dfrac{2}{\phi^2}x^2+\dfrac{1}{\phi^2}}\; dx$
Now let us consider the integrals of the form ,
$\displaystyle F(a,b)= \int\limits_0^\infty \dfrac{x^2+1}{(x^2+a^2)(x^2+b^2)}\; dx$
As we can see that $\displaystyle \dfrac{x^2+1}{(x^2+a^2)(x^2+b^2)} = \dfrac{a^2-1}{(x^2+a^2)(a^2-b^2)}+\dfrac{1-b^2}{(x^2+b^2)(a^2-b^2)}$
Integrating and simplifying we have,
$\displaystyle F(a,b)=\pi\left(\dfrac{1+ab}{2ab(a+b)}\right)$
Now that $\displaystyle x^4+\dfrac{2}{\phi^2}x^2+\dfrac{1}{\phi^2} = \left(x^2+\dfrac{1}{\phi^2}\left(1+i\sqrt{\phi}\right)\right)\left(x^2+\dfrac{1}{\phi^2}\left(1-i\sqrt{\phi}\right)\right)$
We can easily see that our integral equals $\displaystyle I =\dfrac{1}{\phi+1}F\left(\sqrt{\dfrac{1}{\phi^2}\left(1+i\sqrt{\phi}\right)},\sqrt{\dfrac{1}{\phi^2}\left(1-i\sqrt{\phi}\right)}\right)$
Putting the value in we can see that $I=\dfrac{\pi}{2\sqrt{2}}$
The simplification becomes much easier using $\phi^2=\phi+1$ and I've checked it on paper but since it's too much nested so I'm avoiding it in latex. 
