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: On the path of Aditya Narayan Sharma.
Define for $a\geq 0$, 
$\displaystyle F(a)=\int_0^1 \dfrac{1}{(1+ax^4)\sqrt{1-x^2}}dx$
Perform the change of variable $y=\sin x$,
$\displaystyle F(a)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+a(\sin x)^4}dx$
$(\sin x)^2=\dfrac{1}{1+\dfrac{1}{(\tan x)^2}}$
Perform the change of variable $y=\tan x$,
$\begin{align}\displaystyle F(a)&=\int_0^{+\infty} \frac{1}{(1+x^2)\left(1+a\left(\tfrac{1}{1+\frac{1}{x^2}}\right)^2\right)}dx\\
&=\int_0^{+\infty} \dfrac{1+x^2}{(1+a)x^4+2x^2+1}dx\\
\end{align}$
Perform the change of variable $y=x\sqrt[4]{1+a}$,
$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{1+\tfrac{x^2}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$
Perform the change of variable $y=\dfrac{1}{x}$,
$\displaystyle F(a)=\dfrac{1}{\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+\tfrac{1}{\sqrt{1+a}}}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx$
Therefore,
$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{x^2+1}{x^4+\tfrac{2}{\sqrt{1+a}}x^2+1}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_0^{+\infty} \dfrac{\tfrac{1}{x^2}+1}{x^2+\tfrac{1}{x^2}+\tfrac{2}{\sqrt{1+a}}}dx\\
\end{align}$
Perform the change of variable $y=x-\dfrac{1}{x}$,
$\begin{align}
\displaystyle F(a)&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\int_{-\infty}^{+\infty} \dfrac{1}{x^2+2+\tfrac{2}{\sqrt{1+a}}}dx\\
&=\frac{1+\frac{1}{\sqrt{1+a}}}{2\sqrt[4]{1+a}}\left[\dfrac{\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\arctan\left(\dfrac{x\sqrt{1+a}}{\sqrt{2}\sqrt{1+a+\sqrt{1+a}}}\right)\right]_{-\infty}^{+\infty}\\
&=\frac{\sqrt{1+\frac{1}{\sqrt{1+a}}}}{2\sqrt{2}\sqrt[4]{1+a}}\pi\\
&=\boxed{\frac{\sqrt{1+\sqrt{1+a}}}{2\sqrt{2}\sqrt{1+a}}\pi}
\end{align}$
Since $\sqrt{1+\phi}=\phi$ then,
$\begin{align}\displaystyle F(\phi)&=\frac{\sqrt{1+\sqrt{1+\phi}}}{2\sqrt{2}\sqrt{1+\phi}}\pi\\
&=\frac{\sqrt{1+\phi}}{2\sqrt{2}\phi}\pi\\
&=\dfrac{\phi}{2\sqrt{2}\phi}\pi\\
&=\boxed{\dfrac{1}{2\sqrt{2}}\pi}\\
\end{align}$
A: 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. 
A: For $x>0$,
Let,
$\displaystyle F(x)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+x(\sin t)^4}dt$
$\displaystyle H(x)=\int_0^{\tfrac{\pi}{2}} \ln\left(1+x(\sin t)^4\right)dt$
Therefore,
$xH^{\prime} (x)+F(x)=\dfrac{\pi}{2}$
That is,
$\boxed{F(x)=\dfrac{\pi}{2}-xH^{\prime}(x)}$
Using Olivier Oloa's formula, https://math.stackexchange.com/q/873905,
$\displaystyle \boxed{F(x)=\frac{\sqrt{1+\sqrt{1+x}}}{2\sqrt{2}\sqrt{1+x}}\pi}$
