<s>[INST] 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?
 [/INST] \begin{aligned}
I&= \int_{0}^{\frac{\pi}{2}} \frac{x \sin x \cos x}{1-\frac{\sin ^{2} 2 x}{2}} d x\\&=\int_{0}^{\frac{\pi}{2}} \frac{x \sin 2 x}{1+\cos ^{2} 2 x} d x
\\&= \int_{0}^{\frac{\pi}{2}} \frac{\left(\frac{\pi}{2}-x\right) \sin 2 x}{1+\cos ^{2} 2 x} d x \quad \text{ (By }x \mapsto  \frac{\pi}{2}-x.)\\ &=\frac{\pi}{4} \int_{0}^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^{2} 2 x} d x \\
&=-\frac{\pi}{8} \cdot\left[\tan ^{-1}(\cos 2 x)\right]_{0}^{\frac{\pi}{2}} \\
&=-\frac{\pi}{8}\left(-\frac{\pi}{4}-\frac{\pi}{4}\right) \\
&=\frac{\pi^{2}}{16} .
\end{aligned}
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