Putnam definite integral evaluation $\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$ 
Evaluate $$\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$ Source : Putnam

By the property $\displaystyle \int_0^af(x)\,dx=\int_0^af(a-x)\,dx$:
$$=\int_0^{\pi/2}\frac{(\pi/2-x)\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{2}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4 x+\cos^4 x}dx-\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx$$
$$\Longleftrightarrow\int_0^{\pi/2}\frac{x\sin x\cos x}{\sin^4 x+\cos^4 x}dx=\frac{\pi}{4}\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx$$
Now I'm stuck. WolframAlpha says the indefinite integral of $\dfrac{\sin x\cos x}{\sin^4 x+\cos^4x}$ evaluates nicely to $-\frac12\arctan(\cos(2x))$.
I already factored $\sin^4 x+\cos^4 x$ into $1-\left(\frac{\sin(2x)}{\sqrt{2}}\right)^2$, but I don't know how to continue.. I suggest a substitution $u=\frac{\sin(2x)}{\sqrt{2}}$?
Could someone provide me a hint, or maybe an easier method I can refer to in the future?

$1 - \dfrac{\sin^2(2x)}{2} = \dfrac{1+\cos^2(2x)}{2}$, and $\sin x\cos x = \dfrac{\sin (2x)}{2} \Rightarrow \displaystyle \int \dfrac{\sin x\cos x}{\cos^4x+\sin^4x}dx = \displaystyle \int -\dfrac{1}{2}\dfrac{d(\cos(2x))}{1+\cos^2(2x)}dx = -\dfrac{1}{2}\arctan(\cos (2x)) + C$
