The value of the integral $$\frac{48}{\pi^4}\int_0^{\pi}\left(\frac{3\pi x^2}{2} - x^3\right)\frac{\sin x}{1 + \cos^2 x}dx$$ is equal to ______

We need: $$\frac{48}{\pi^4}\int_0^{\pi}\left(\frac{3\pi x^2}{2} - x^3\right)\frac{\sin x}{1 + \cos^2 x}\,dx$$

Since $$\frac{3\pi x^2}{2} - x^3 = x^2\left(\frac{3\pi}{2} - x\right)$$, we set $$I = \int_0^{\pi} x^2\left(\frac{3\pi}{2} - x\right)\frac{\sin x}{1 + \cos^2 x}\,dx$$. Now using the substitution $$x\to \pi - x$$ leaves $$\sin x$$ and $$\cos^2 x$$ unchanged, and transforms the integrand factor as $$(\pi - x)^2\left(\frac{3\pi}{2} - \pi + x\right) = (\pi - x)^2\left(\frac{\pi}{2} + x\right)$$. Thus the same integral can be written as $$I = \int_0^{\pi}(\pi - x)^2\left(\frac{\pi}{2} + x\right)\frac{\sin x}{1 + \cos^2 x}\,dx$$.

Next, adding these two expressions for $$I$$ gives $$2I = \int_0^{\pi} \left[x^2\left(\frac{3\pi}{2} - x\right) + (\pi - x)^2\left(\frac{\pi}{2} + x\right)\right]\frac{\sin x}{1 + \cos^2 x}\,dx$$. Expanding the bracket yields $$x^2\left(\frac{3\pi}{2} - x\right) = \frac{3\pi x^2}{2} - x^3$$ and $$(\pi - x)^2\left(\frac{\pi}{2} + x\right) = (\pi^2 - 2\pi x + x^2)\left(\frac{\pi}{2} + x\right) = \frac{\pi^3}{2} - 2\pi x^2 + \frac{\pi x^2}{2} + x^3 = \frac{\pi^3}{2} - \frac{3\pi x^2}{2} + x^3$$. From this the sum simplifies to $$\frac{3\pi x^2}{2} - x^3 + \frac{\pi^3}{2} - \frac{3\pi x^2}{2} + x^3 = \frac{\pi^3}{2}$$.

Therefore $$2I = \frac{\pi^3}{2}\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\,dx$$. To evaluate the remaining integral let $$u = \cos x$$ so that $$du = -\sin x\,dx$$ and hence $$\int_0^{\pi}\frac{\sin x}{1 + \cos^2 x}\,dx = \int_1^{-1}\frac{-\,du}{1 + u^2} = \int_{-1}^{1}\frac{du}{1 + u^2} = [\tan^{-1}u]_{-1}^{1} = \frac{\pi}{4} - \left(-\frac{\pi}{4}\right) = \frac{\pi}{2}$$. Substituting back gives $$2I = \frac{\pi^3}{2}\cdot \frac{\pi}{2} = \frac{\pi^4}{4}$$, so that $$I = \frac{\pi^4}{8}$$.

Finally, multiplying by the coefficient yields $$\frac{48}{\pi^4}\cdot \frac{\pi^4}{8} = 6$$.

The answer is $$\boxed{6}$$.