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): $$\int_0^\infty e^{-tx}\cos(t)\d t=\frac{x}{x^2+1},\,x\gt0$$
 A: 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}
A: $$\begin{split}
I &= \int_{-\frac \pi 2}^{\frac \pi 2}\sum_{n\geq 0}(-1)^n \sin^{4n}x \,\,dx\\
&= 2\sum_{n\geq 0}(-1)^n \int_{0}^{\frac \pi 2}\sin^{4n}x \,\,dx\\
&= 2\sum_{n\geq 0}(-1)^n \cdot\frac \pi 2 \frac{(4n)!}{2^{4n}((2n)!)^2}\\
&= \pi f\left(-\frac 1 {16}\right)\\
&=\frac \pi 2 \left ( \frac 1 {\sqrt{1-i}} +\frac 1 {\sqrt{1+i}}  \right)\\
&= \frac {\pi\sqrt{1+\sqrt 2}}{2}
\end{split}$$
where we have used, in order:
$$\begin{array}{ll}
\text{(1) Used the geometric series expansion of } \frac 1 {1+\sin^4 x}.\\
\text{(2) Interchanged the integral and sum}.\\
\text{(3) Used the known formula for the Wallis integral}.\\
\text{(4) Set }f(z)=\sum_{n\geq 0}\binom{4n}{2n}z^n=\frac 1 2 \left( \frac 1 {\sqrt{1-4\sqrt z}}+ \frac 1 {\sqrt{1+4\sqrt z}} \right )\\
\end{array}$$
To compute $(4)$, we used the known formula for the binomial series to obtain:
$$g(z) = \sum_{n\geq 0}\binom{2n}{n}z^n=\frac 1 {\sqrt {1-4z}}$$
and then note that $f(z) = \frac 1 2 \left(g(\sqrt z)+g(-\sqrt z)\right)$.
A: 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
$$
\begin{aligned}I&=\int_{0}^{\infty } \frac{\frac{1+\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}+\frac{1}{t^{2}}\right)+\frac{1-\sqrt{2}}{\sqrt{2}}\left(\sqrt{2}-\frac{1}{t^{2}}\right)}{2 t^{2}+2+\frac{1}{t^{2}}} d t\\& =\frac{1+ \sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t-\frac{1}{t}\right)}{\left(\sqrt{2} t-\frac{1}{t}\right)^{2}+2(\sqrt{2}+1)}+\frac{1-\sqrt{2}}{\sqrt{2}} \int_{0}^{\infty} \frac{d\left(\sqrt{2} t+\frac{1}{t}\right)}{\left(\sqrt{2} t+\frac{1}{t}\right)^{2}-2(\sqrt{2}-1)}\\& =\frac{1+\sqrt{2}}{2 \sqrt{\sqrt{2}+1}}\left[\tan ^{-1} \frac{\left(\sqrt{2} t-\frac{1}{t}\right)}{\sqrt{2(\sqrt{2}+1)}}\right]_{0}^{\infty}+\frac{1-\sqrt{2}}{4 \sqrt{\sqrt{2}-1}} \ln \left|\frac{\left.\sqrt{2} t+\frac{1}{t}-\sqrt{2(\sqrt{2}-1}\right) }{\sqrt{2} t+\frac{1}{t}+\sqrt{2}(\sqrt{2}-1}\right|_0^{\infty}\\&= \frac{\pi}{2} \sqrt{1+\sqrt{2}}\end{aligned} $$
A: Working the antiderivative
Using your first step
$$\int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du$$
$$\left(u^2+1\right)^2+u^4=2 \left(u^2-u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)
   \left(u^2+u\sqrt{\sqrt{2}-1} +\frac{1}{\sqrt{2}}\right)$$ Partial fraction decomposition
$$ \frac{u^2+1}{\left(u^2+1\right)^2+u^4}=$$
$$\frac 1 {2 \sqrt{2 \left(\sqrt{2}-1\right)} }\Bigg[\frac {\left(\sqrt{2}-2\right) u+2
   \sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }+\frac {\left(2-\sqrt{2}\right) u+2
   \sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}  }\Bigg]$$
$$4\int \frac {\left(\sqrt{2}-2\right) u+2
   \sqrt{\sqrt{2}-1} } {2 u^2-2 \sqrt{\sqrt{2}-1} u+\sqrt{2} }\,du=$$ $$\left(\sqrt{2}-2\right) \log \left(2 u^2-2 \sqrt{\sqrt{2}-1}
   u+\sqrt{2}\right)+2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan
   ^{-1}\left(\frac{2 u-\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)$$
$$4\int\frac {\left(2-\sqrt{2}\right) u+2
   \sqrt{\sqrt{2}-1}} {2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}  }\,du$$
$$2 \sqrt{3-2 \sqrt{2}} \left(2+\sqrt{2}\right) \tan ^{-1}\left(\frac{2
   u+\sqrt{\sqrt{2}-1}}{\sqrt{1+\sqrt{2}}}\right)-\left(\sqrt{2}-2\right) \log
   \left(2 u^2+2 \sqrt{\sqrt{2}-1} u+\sqrt{2}\right)$$
$$\color{red}{8 \sqrt{2 \left(\sqrt{2}-1\right)} \int \frac{u^2+1}{\left(u^2+1\right)^2+u^4} \,du=}$$
$$\color{red}{-2 \sqrt{2} \tan ^{-1}\left(\frac{2 \sqrt{\sqrt{2}-1} u}{2 \left(\sqrt{2}-1\right)
   u^2+\sqrt{2}-2}\right)-\left(2-\sqrt{2}\right) \log \left(\frac{2 u
   \left(u-\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}{2 u
   \left(u+\sqrt{\sqrt{2}-1}\right)+\sqrt{2}}\right)}$$ gives the desired result.
