<s>[INST] How to integrate $\int \frac{1}{\sin^4x + \cos^4 x} \,dx$? 
How to integrate 
  $$\int \frac{1}{\sin^4x + \cos^4 x} \,dx$$ 

I tried the following approach:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{\sin^4x + (1-\sin^2x)^2} \,dx = \int \frac{1}{\sin^4x + 1- 2\sin^2x + \sin^4x} \,dx \\
= \frac{1}{2}\int \frac{1}{\sin^4x - \sin^2x + \frac{1}{2}} \,dx = \frac{1}{2}\int \frac{1}{(\sin^2x - \frac{1}{2})^2 + \frac{1}{4}} \,dx$$
The substitution $t = \tan\frac{x}{2}$ yields 4th degree polynomials and a $\sin$ substitution would produce polynomials and expressions with square roots while Wolfram Alpha's solution doesn't look that complicated. Another approach:
$\sin^4x + \cos^4 x = (\sin^2 x + \cos^2x)(\sin^2 x + \cos^2 x) - 2\sin^2 x\cos^2 x = 1 - 2\sin^2 x\cos^2 x  = (1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)$
 and then I tried substituting: $t = \sin x \cos x$ and got 
$$\int\frac{t\,dt}{2(1-2t^2)\sqrt{1-4t^2}}$$
Another way would maybe be to make two integrals:
$$\int \frac{1}{\sin^4x + \cos^4 x} \,dx = \int \frac{1}{(1-\sqrt2\sin x \cos x)(1+\sqrt2\sin x \cos x)} \,dx = \\ \frac{1}{2}\int \frac{1}{1-\sqrt2\sin x \cos x} \,dx +  \frac{1}{2}\int\frac{1}{1+\sqrt2\sin x \cos x} \,dx$$ 
... and again I tried $t = \tan\frac{x}{2}$ (4th degree polynomial) and $t=\sqrt2 \sin x \cos x$ and I get $\frac{\sqrt 2}{2} \int \frac{\,dt}{(1-t)\sqrt{1-2t^2}}$ for the first one.
Any hints?
 [/INST] Convert the exponential powers to multiple angles. From deMoivre's theorem, with $n\in\mathbb{N}$:
$$
\begin{align}
\left( e^{i \theta} \right)^{n} &= e^{i n\theta} \\
\left( \cos \theta + i \sin \theta \right)^{n} &= \cos n\theta + i \sin n\theta
\end{align}
$$
These intermediate formulas may help:
$$
\begin{align}
  \cos 2\theta &= \cos^{2} \theta - \sin^{2} \theta \\
  \sin 2\theta &= 2 \cos \theta \sin 2 \theta \\
\end{align}
$$
Reduce the denominator
$$
\sin ^4(x)+\cos ^4(x) = \frac{1}{4} (\cos (4 x)+3)
$$
The primitive is
$$
\int \frac{1}{\cos^{4}x + \sin^{4}x} \, dx =
\int \frac{1}{\cos (4 x)+3} \, dx = 
\left(4 \sqrt{2}\right)^{-1}\arctan \left(\frac{\tan (2 x)}{\sqrt{2}}\right)
$$
Here is a look at the integrand:

</s>