Integrate $\int \frac{\sin x \cos x}{\sin^4x + \cos^4x} \,dx$ 
Integrate $$\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}dx$$

I solved the question by using the identity $\cos^4(x)+\sin^4(x) = \frac{1}{4}(\cos4x+3)$ and the substitution $u=\cos4x +3$, which turned it into a relatively familiar integral (see my answer below). 
However, I'm pretty sure there are easier ways I am missing, so please feel free to post alternative answers. 
There is a similar question here. 
Problem Source: James Stewart Calculus, 6E

$u = \cos 2t \implies du = -2\sin 2t dt$
$$\begin{align}\dfrac{-1}2\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}\dfrac{du}{\sin 2t} =& \dfrac{-1}4\int \frac{1}{\sin^4x + \cos^4x}du &\\=& \dfrac{-1}2\int \frac{1}{(\cos^2x + \sin^2x)^2 +(\cos^2x - \sin^2x)^2 } &\\=& \dfrac{-1}2\int \frac{1}{1+u^2 }du = \dfrac{-1}{2}\tan^{-1}(\cos 2x) + C\end{align}$$
