<s>[INST] 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
 [/INST] $$ \frac{\sin x\cos x}{\sin^4 x + \cos^4 x} = \frac{\sin 2x}{2(1 - 2\sin^2x \cos^2 x)} = \frac{\sin 2x}{2 - (1-\cos 2x)(1 + \cos 2x)} $$
Substitute $u = \cos 2x$ to get
$$ -\frac{1}{2}\int\frac{du}{1+u^2} = -\frac{1}{2}\arctan u = \color{blue}{-\frac{1}{2}\arctan (\cos 2x)} $$
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