Solutions:
\( I = \int_{-\pi/2}^{\pi/2} \left( \frac{x^2 \cos x}{1 + \pi^x} + \frac{1 + \sin^2 x}{1 + e^{\sin x^{2023}}} \right) dx \)
Let \( I = I_1 + I_2 \)
\( I_1 = \int_{-\pi/2}^{\pi/2} \frac{x^2 \cos x}{1 + \pi^x} dx \quad \ldots \text{(i)} \)
Replace \( x \) by \( -x \)
\( I_1 = \int_{-\pi/2}^{\pi/2} \frac{\pi^x \left( x^2 \cos x \right)}{1 + \pi^x} dx \quad \ldots \text{(ii)} \)
Add (i) and (ii)
\( 2 I_1 = \int_{-\pi/2}^{\pi/2} x^2 \cos x \, dx \)
\( I_1 = \int_{-\pi/2}^{\pi/2} x^2 \cos x \, dx \quad \left[ \text{If } f(-x)=f(x) \text{ then } \int_{-a}^{a} f(x)\,dx = 2 \int_{0}^{a} f(x)\,dx \right] \)
\( I_1 = x^2 \int \cos x \, dx - \int (2x) \int \cos x \, dx \)
\( = x^2 \sin x - 2 \int x \sin x \, dx \)
\( = x^2 \sin x - 2[ x(-\cos x) - (-\sin x) ] \)
\( = [ x^2 \sin x + 2x \cos x - 2 \sin x ]_{0}^{\pi/2} = \frac{\pi^2}{4} - 2 \)
\( I_2 = \int_{-\pi/2}^{\pi/2} \frac{1 + \sin^2 x}{1 + e^{\sin x^{2023}}} dx \quad \ldots \text{(iii)} \)
Replace \( x \) by \( -x \)
\( I_2 = \int_{-\pi/2}^{\pi/2} \frac{e^{\sin x^{2023}} (1 + \sin^2 x)}{1 + e^{\sin x^{2023}}} dx \quad \ldots \text{(iv)} \)
Add (iii) and (iv)
\( 2 I_2 = \int_{-\pi/2}^{\pi/2} (1 + \sin^2 x) dx \)

\( I_2 = \left[ x \right]_{0}^{\pi/2} + \int_{0}^{\pi/2} \left[ \frac{1 - \cos 2x}{2} \right] dx \)

\( I = I_1 + I_2 = \frac{\pi^2}{4} - 2 + \frac{3\pi}{4} = \frac{\pi}{4} (\pi + 3) - 2 \)
a=3

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