$\newcommand{\d}{\,\mathrm{d}}$Last night, I evaluated the following integral: $$\begin{align}I:&=\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\d x\\&=\int_{-1}^1\frac{1}{(1+x^4)\sqrt{1-x^2}}\d x\\&=\frac{\pi}{2^{3/4}}
(\sin(\pi/8)+\cos(\pi/8))\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Using a "double keyhole" (as I phrase it) contour method involving a management of branch cuts and residues at infinity, here. Although I was happy to have succeeded in this, I wondered afterwards if I would have had any hope of evaluating $I$ with real analytic technique only.
The challenge:
Evaluate $I$ without use of complex analysis or even of complex arithmetic (e.g. for partial fraction decompositions involving $i$)
I posed this to some friends and they came up with the following method which I wanted to share with MSE:
$$\begin{align}I&\overset{x\mapsto\tan x}{=}\int_{-\infty}^\infty\frac{1+x^2}{(1+x^2)^2+x^4}\d x\\&\overset{x\mapsto1/x}{=}2\int_0^\infty\frac{1+x^2}{(1+x^2)^2+1}\d x\\&=2\int_0^\infty\int_0^\infty e^{-t(1+x^2)}\cos(t)\d t\d x\quad\text{Repr. with IBP}\\&=\sqrt{\pi}\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\end{align}$$
$$\begin{align}J:&=\int_0^\infty\frac{e^{-t}\cos(t)}{\sqrt{t}}\d t\\J^2&=\int_0^\infty\int_0^\infty\frac{e^{-(t+x)}\cos(t)\cos(x)}{\sqrt{tx}}\d t\d x\\&\overset{x\mapsto tx}{=}\int_0^\infty\int_0^\infty\frac{e^{-t(1+x)}\cos(t)\cos(tx)}{\sqrt{x}}\d x\d t\\&=\frac{1}{2}\int_0^\infty\frac{1}{\sqrt{x}}\cdot\frac{1+x+x^2}{(1+x)(1+x^2)}\d x\\&\overset{x\mapsto x^2}{=}\frac{1}{2}\int_0^\infty\left(\frac{1+x^2}{1+x^4}+\frac{1}{1+x^2}\right)\d x\\&=\frac{1}{2}\left[\frac{\pi}{4}\csc\left(\frac{\pi}{4}\right)+\frac{\pi}{4}\csc\left(\frac{3\pi}{4}\right)+\frac{\pi}{2}\right]\\&=\frac{\pi}{4}(1+\sqrt{2})\end{align}$$
Referencing this answer by Sangchul.
We conclude:
$$\begin{align}I&=\sqrt{\pi}\cdot\sqrt{J^2}\\&=\sqrt{\pi}\cdot\sqrt{\frac{\pi}{4}(1+\sqrt{2})}\\&=\frac{\pi}{2}\sqrt{1+\sqrt{2}}\end{align}$$
Among those who helped me, who use MSE, I credit @TheSimpliFire and @KStarGamer who are much better at real integration than I am!
My question is less of a question and more of a request for a list – a list of other, purely real, methods to attack this integral. I hope the outcome of this will be an interesting selection of advanced integration techniques that I and others can learn from.
Note 1: I am aware of this posting by Quanto but it uses complex numbers.
Note 2: You must expand the cosine product as a sum of cosines and use the same integral representation (which is classically gotten from complex arithmetic but can be done with integration by parts): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$
Best Answer
Alternatively \begin{align} I=& \int_{0}^{\pi/2}\frac{1}{1+\sin^4x} \overset{t=\sqrt[4]2\tan x} {dx} + \int_{0}^{\pi/2}\frac{1}{1+\cos ^4x} \overset{\sqrt[4]2 t=\tan x} {dx }\\ = &\ \frac{{1+\sqrt2}}{2^{3/4}}\int_0^\infty\frac{1+t^2}{t^4+\sqrt2t^2+1}dt = \frac{{1+\sqrt2}}{2^{3/4}}\int_0^\infty\frac{d(t-\frac1t)}{(t-\frac1t)^2+(2+\sqrt2)}\\ =& \ \frac\pi2 \sqrt{1+\sqrt2}\\ \end{align}