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

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 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.