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?

Here is one line proof
$$\int_0^{\pi/2}\frac{\sin x\cos x}{\sin^4x+\cos^4x}dx=\int_0^{\pi/2}\frac{\tan(x) (\tan(x))'}{\tan^4(x)+1}\ dx=\int_0^{\infty} \frac{x}{x^4+1}\ dx=\left[\frac{\arctan(x^2)}{2}\right]_0^{\infty}=\frac{\pi}{4}$$
Q.E.D.
