$$\frac{120}{\pi^3}\int_{0}^{\pi}\frac{x^2\sin x\cos x}{\sin^4 x + \cos^4 x}dx$$ is equal to

Use King's Property ($$x \to \pi - x$$):

$$I = \frac{120}{\pi^3} \int_0^\pi \frac{(\pi-x)^2 \sin x (-\cos x)}{\sin^4 x + \cos^4 x} dx$$.

Due to the $$\cos x$$ term changing sign, this typically requires splitting the integral at $$\pi/2$$. Using the symmetry of the denominator and the $$x^2$$ term, the integral evaluates such that the $$\pi$$ terms cancel out.

The core integral $$\int \frac{\sin x \cos x}{\sin^4 x + \cos^4 x} dx$$ is solved by dividing by $$\cos^4 x$$ and using substitution $$u = \tan^2 x$$.

The final calculation results in 15