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
 A: $$ \frac{\sin x\cos x}{\sin^4 x + \cos^4 x} = \frac{\sin 2x}{2(1 - 2\sin^2x \cos^2 x)} = \frac{\sin 2x}{2 - (1-\cos 2x)(1 + \cos 2x)} $$
Substitute $u = \cos 2x$ to get
$$ -\frac{1}{2}\int\frac{du}{1+u^2} = -\frac{1}{2}\arctan u = \color{blue}{-\frac{1}{2}\arctan (\cos 2x)} $$
A: First, some preliminary manipulation. $$\frac{\sin x \cos x}{\sin^4x + \cos^4x} = \frac{\sin x \cos x}{(1-\cos^2x)^2 + \cos^4x}\\=\frac{\sin x \cos x}{2\cos^4x -2\cos^2x+1}=\frac{4\sin x \cos x}{8\cos^4x -8\cos^2x+1 + 3}\\=\frac{4\sin x \cos x}{\cos4x+ 3}$$
The last step uses the quadruple angle formula for cosine. Now $\sin 2x = 2\sin x \cos x$, using this twice yields:$$\frac{4\sin x \cos x}{\cos4x+ 3}=\frac{2\sin 2x }{\cos4x+ 3} =\frac{2\sin 2x \cos 2x}{(\cos4x+ 3)\cos 2x} = \frac{\sin 4x }{(\cos4x+ 3)\cos 2x} $$
We can now use the half-angle formula for cosine, which is $\cos\frac{x}{2}=\sqrt{\frac{1+\cos x}{2}}$. $$\frac{\sin 4x }{(\cos4x+ 3)\cos 2x}=\frac{\sin 4x }{(\cos4x+ 3)\sqrt{\frac{1+\cos 4x}{2}}} \\=\frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}}$$
The time is ripe to substitute with $u=\cos 4x +3$. Then $du = -4\sin 4x \ dx$ and $$\int \frac{\sqrt{2} \times \sin 4x }{(\cos4x+ 3)\sqrt{\cos 4x + 3 - 2}} \ dx\\ = \frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du$$
To finish this relatively simple integral, I did another substitution, this time with $t=\sqrt{u-2}$ and $du =2t \ dt$. $$\frac{\sqrt{2}}{-4}\int \frac{1}{u\sqrt{u - 2}} \ du = \frac{\sqrt{2}}{-4}\int \frac{2t}{ut} \ dt\\=\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt$$
This is an integral that involves $\arctan$:$$\frac{\sqrt{2}}{-2}\int \frac{1}{t^2 +2} \ dt=-\frac{1}{2}\arctan(\frac{t}{\sqrt{2}})=-\frac{1}{2}\arctan(\frac{\sqrt{u-2}}{\sqrt{2}})\\=-\frac{1}{2}\arctan(\sqrt{\frac{\cos 4x +1}{2}})=-\frac{1}{2} \arctan(\cos 2x)$$
Checking with wolframalpha, this differentiates to the correct result.
A: Hint:
Divide numerator & denominator by  $\cos^4x$
and set $\tan^2x=u$
Alternatively, the divisor $=\sin^4x$
and the substitution $=\cot^2x=v$
A: $u = \cos 2t \implies du = -2\sin 2t dt$
$$\begin{align}\dfrac{-1}2\int \frac{\sin x \cos x}{\sin^4x + \cos^4x}\dfrac{du}{\sin 2t} =& \dfrac{-1}4\int \frac{1}{\sin^4x + \cos^4x}du &\\=& \dfrac{-1}2\int \frac{1}{(\cos^2x + \sin^2x)^2 +(\cos^2x - \sin^2x)^2 } &\\=& \dfrac{-1}2\int \frac{1}{1+u^2 }du = \dfrac{-1}{2}\tan^{-1}(\cos 2x) + C\end{align}$$
A: $$
\int \frac{\sin x \cos x}{\sin ^{4} x+\cos ^{4} x} d x=\int \frac{\frac{\sin 2 x}{2}}{1-\frac{\sin ^{2} 2 x}{2}} d x =\int \frac{\sin 2 x}{2-\sin ^{2} 2 x} d x =-\frac{1}{2} \int \frac{d(\cos (2 x))}{1+\cos ^{2}(2 x)} \\\boxed{=-\frac{1}{2} \arctan (\cos 2 x)+C}
$$
