What purely real analytic techniques are there to evaluate $\int_{-\pi/2}^{\pi/2}\frac{1}{1+\sin^4(x)}\,\mathrm{d}x$? $\newcommand{\d}{\,\mathrm{d}}$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): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$

Using symmetry of sine and letting $t=\tan x$ yields
$$
\begin{array}{l}
\displaystyle I=:\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1}{1+\sin ^{4} x} d x=2 \int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\sec ^{4} x+\tan ^{4} x} d x 
=2 \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+t^{2}\right)^{2}+t^{4}} d t
\end{array}
$$
Dividing both numerator and denominator by $t^2$ yields
$$
\begin{aligned}I&=\int_{0}^{\infty } \frac{\frac{1+\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}+\frac{1}{t^{2}}\right)+\frac{1-\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}-\frac{1}{t^{2}}\right)}{2 t^{2}+2+\frac{1}{t^{2}}} d t\\& =\frac{1+ \sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t-\frac{1}{t}\right)}{\left(\sqrt{2} t-\frac{1}{t}\right)^{2}+2(\sqrt{2}+1)}+\frac{1-\sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t+\frac{1}{t}\right)}{\left(\sqrt{2} t+\frac{1}{t}\right)^{2}-2(\sqrt{2}-1)}\\& =\frac{1+\sqrt{2}}{2 \sqrt{\sqrt{2}+1}}\left[\tan ^{-1} \frac{\left(\sqrt{2} t-\frac{1}{t}\right)}{\sqrt{2(\sqrt{2}+1)}}\right]_{0}^{\infty}+\frac{1-\sqrt{2}}{4 \sqrt{\sqrt{2}-1}} \ln \left|\frac{\left.\sqrt{2} t+\frac{1}{t}-\sqrt{2(\sqrt{2}-1}\right) }{\sqrt{2} t+\frac{1}{t}+\sqrt{2}(\sqrt{2}-1}\right|_0^{\infty}\\&= \frac{\pi}{2} \sqrt{1+\sqrt{2}}\end{aligned} $$
