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

$I=\frac{48}{{\pi }^4}{\int }_0^{\pi } x^2\left(\frac{3\pi }{2}-x\right)\frac{\sin x}{1+{\cos }^{2}x}dx$-------(1)

$$\begin{gathered} I=\frac{48}{{\pi }^4}{\int }_0^{\pi } (\pi -x{)}^2\left(\frac{\pi }{2}+x\right)\frac{\sin x}{1+{\cos }^{2}x}dx----(2\left) \\ \right(1)+(2)\Rightarrow \\ \mathrm{I}=\frac{12}{{\pi }^3}{\int }_0^{\pi } \frac{\sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}\left[{\pi }^2+(\pi -2)\cdot \mathrm{x}\cdot (\pi -2\mathrm{x})\right]\mathrm{dx}\dots (3) \end{gathered}$$

$$\begin{gathered} I=\frac{12}{{\pi }^3}{\int }_0^{\pi } \frac{\sin x}{1+{\cos }^{2}x}\left[{\pi }^2+(\pi -2)(\pi -x)(2x-\pi )\right]dx-----\left(4\right) \\ \left(3\right)+\left(4\right)\Rightarrow \\ \mathrm{I}=\frac{6}{{\pi }^2}{\int }_0^{\pi } \frac{\sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}[2\pi +(\pi -2\left)\right(\pi -2\mathrm{x}\left)\right]\mathrm{dx}---\left(5\right) \end{gathered}$$

$$\begin{gathered} \mathrm{I}=\frac{6}{{\pi }^2}{\int }_0^{\pi } \frac{\sin \mathrm{x}}{1+{\cos }^2\mathrm{x}}[2\pi +(\pi -2\left)\right(2\mathrm{x}-\pi \left)\right]\mathrm{dx}\dots \left(6\right) \\ \left(5\right)+\left(6\right)\Rightarrow \\ I=\frac{12}{\pi }{\int }_0^{\pi } \frac{\sin x}{1+{\cos }^{2}x}dx \\ \text{ Let }\cos \mathrm{x}=\mathrm{t}\Rightarrow \sin \mathrm{xdx}=-\mathrm{dt} \\ \mathrm{I}=\frac{12}{\mathrm{\pi }}{\int }_1^{-1} \frac{-\mathrm{dt}}{1+{\mathrm{t}}^2}=6 \end{gathered}$$