Let $I=\frac{48}{\pi^4}\int_0^{\pi}\left(\frac{3\pi x^2}{2}-x^3\right)\frac{\sin x}{1+\cos^2 x}\,dx.$

We will first simplify the polynomial factor, then evaluate the definite integral.

1. Factor the polynomial

Observe that $\frac{3\pi x^2}{2}-x^3=x^2\left(\frac{3\pi}{2}-x\right).$

A more useful symmetry form is obtained by shifting around $x=\frac{\pi}{2}$.

Let $x=\frac{\pi}{2}+t.$ Then $\frac{3\pi x^2}{2}-x^3=x^2\left(\frac{3\pi}{2}-x\right).$

But an even cleaner approach is to use the symmetry of $f(x)=\frac{\sin x}{1+\cos^2 x}.$ Since $f(\pi-x)=\frac{\sin(\pi-x)}{1+\cos^2(\pi-x)}=\frac{\sin x}{1+\cos^2 x}=f(x),$ we can use the identity $\int_0^{\pi}P(x)f(x)\,dx=\int_0^{\pi}\frac{P(x)+P(\pi-x)}{2}f(x)\,dx.$

Here $P(x)=\frac{3\pi x^2}{2}-x^3.$ Compute $P(\pi-x)$: $P(\pi-x)=\frac{3\pi}{2}(\pi-x)^2-(\pi-x)^3.$

Now simplify: $\frac{P(x)+P(\pi-x)}{2}.$

First expand: $(\pi-x)^2=\pi^2-2\pi x+x^2,$ $(\pi-x)^3=\pi^3-3\pi^2x+3\pi x^2-x^3.$

So $P(\pi-x)=\frac{3\pi}{2}(\pi^2-2\pi x+x^2)-\left(\pi^3-3\pi^2x+3\pi x^2-x^3\right).$

This becomes $P(\pi-x)=\frac{3\pi^3}{2}-3\pi^2x+\frac{3\pi x^2}{2}-\pi^3+3\pi^2x-3\pi x^2+x^3,$ $P(\pi-x)=\frac{\pi^3}{2}-\frac{3\pi x^2}{2}+x^3.$

Hence $P(x)+P(\pi-x)=\left(\frac{3\pi x^2}{2}-x^3\right)+\left(\frac{\pi^3}{2}-\frac{3\pi x^2}{2}+x^3\right)=\frac{\pi^3}{2}.$

Therefore, $\frac{P(x)+P(\pi-x)}{2}=\frac{\pi^3}{4}.$

So the inner integral becomes $\int_0^{\pi}P(x)f(x)\,dx=\frac{\pi^3}{4}\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx.$

2. Evaluate the remaining trigonometric integral

Let $J=\int_0^{\pi}\frac{\sin x}{1+\cos^2 x}\,dx.$

Use substitution: $u=\cos x,\qquad du=-\sin x\,dx.$

When $x=0$, $u=1$; when $x=\pi$, $u=-1$. Thus $J=\int_1^{-1}\frac{-du}{1+u^2}=\int_{-1}^1\frac{du}{1+u^2}.$

Therefore, $J=\left[\tan^{-1}u\right]_{-1}^{1}=\frac{\pi}{4}-\left(-\frac{\pi}{4}\right)=\frac{\pi}{2}.$

3. Compute the original integral

So