Given,

${\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(\frac{x^2\cos x}{1+{\pi }^x}+\frac{1+{\sin }^{2}x}{1+e^{{\sin }^{2023}x}}\right)dx=\frac{\pi }{4}\left(\pi +\alpha \right)-4$

Now let, $I_1={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(\frac{x^2\cos x}{1+{\pi }^x}\right)dx .......\left(i\right)$

Now using the property ${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f\left(a+b-x\right)dx$ we get,

$\Rightarrow I_1={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(\frac{x^2\cos x}{1+{\pi }^{-x}}\right)dx ........\left(ii\right)$

Now adding above equations we get,

$\Rightarrow 2I_1=2{\int }_0^{\frac{\pi }{2}}\left(x^2\cos x\right)dx$ $\left\{\text{as} {\int }_{-a}^{a}f\left(x\right)dx=2{\int }_0^{a}f\left(x\right)dx, \text{if} f\left(x\right) \text{is even}\right\}$

$\Rightarrow I_1={\left[x^2\left(\sin x\right)\right]}_0^{\frac{\pi }{2}}-{\int }_0^{\frac{\pi }{2}}2x·\sin xdx$

$\Rightarrow I_1=\frac{{\pi }^2}{4}-\left[{\left[-2x\cos x\right]}_0^{\frac{\pi }{2}}+{\int }_0^{\frac{\pi }{2}}2\cos xdx\right]$

$\Rightarrow I_1=\frac{{\pi }^2}{4}-2$

Similarly solving, $I_2={\int }_{\frac{-\pi }{2}}^{\frac{\pi }{2}}\left(\frac{1+{\sin }^{2}x}{1+e^{{\sin }^{2023}x}}\right)dx$

$\Rightarrow I_2={\int }_0^{\frac{\pi }{2}}\left(1+{\sin }^{2}x\right)dx$

$\Rightarrow I_2={\int }_0^{\frac{\pi }{2}}\left(1+\frac{1-\cos 2x}{2}\right)dx$

$\Rightarrow I_2=\frac{1}{2}{\int }_0^{\frac{\pi }{2}}\left(3-\cos 2x\right)dx$

$\Rightarrow I_2=\frac{1}{2}{\left[3x-\frac{\sin 2x}{2}\right]}_0^{\frac{\pi }{2}}$

$\Rightarrow I_2=\frac{3\pi }{4}$

$\Rightarrow I_1+I_2=\frac{{\pi }^2}{4}-2+\frac{3\pi }{4}$

$\Rightarrow \frac{\pi }{4}\left(\pi +3\right)-2=\frac{\pi }{4}\left(\pi +\alpha \right)-2$

$\Rightarrow \alpha =3$