Linearity Property

Let

$$I= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \frac{x^2\cos x}{1+\pi^x} + \frac{1+\sin^2x}{1+e^{\sin x^{2023}}} \right)\,dx$$

Write

$$I=I_1+I_2$$

where

$$I_1= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{x^2\cos x}{1+\pi^x}\,dx$$ and $$I_2= \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{1+\sin^2x}{1+e^{\sin x^{2023}}}\,dx$$

Using

$$\int_{-a}^{a}f(x)\,dx = \int_{0}^{a}[f(x)+f(-x)]\,dx,$$

we get

$$I_1=\int_{0}^{\frac{\pi}{2}}\left(\frac{x^2\cos x}{1+\pi^x}+\frac{x^2\cos x{1+\pi^{-x}}\right)\,dx$$

Since

$$\frac1{1+\pi^x}+\frac1{1+\pi^{-x}}=1,$$

therefore

$$I_1=\int_{0}^{\frac{\pi}{2}}x^2\cos x\,dx$$

Integrating by parts,

$$I_1=\left[x^2\sin x+2x\cos x-2\sin x\right]_0^{\frac{\pi}{2}}$$

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

Now,

$$I_2= \int_{0}^{\frac{\pi}{2}} (1+\sin^2x)\,dx$$

because

$$\frac1{1+e^{\sin x^{2023}}} + \frac1{1+e^{-\sin x^{2023}}}=1$$

Hence,

$$I_2= \frac{\pi}{2}+\frac{\pi}{4}$$

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

Therefore,

$$I= \frac{\pi^2}{4}-2+\frac{3\pi}{4}$$

$$= \frac{\pi}{4}(\pi+3)-2$$

Comparing with

$$\frac{\pi}{4}(\pi+a)-2,$$

we get

$$a=3$$

$$\int\frac1{\sqrt{a^2-b^2x^2}}\,dx = \frac1b\sin^{-1}\left(\frac{bx}{a}\right),$$

we get

$$I= 48\cdot\frac12 \left[ \sin^{-1}\left(\frac{2x}{3}\right) \right]_{\frac{3\sqrt2}{4}}^{\frac{3\sqrt3}{4}}$$

$$= 24\left[ \sin^{-1}\left(\frac{\sqrt3}{2}\right) -\sin^{-1}\left(\frac1{\sqrt2}\right) \right]$$

$$= 24\left[ \frac{\pi}{3}-\frac{\pi}{4} \right]$$

$$= 24\left(\frac{\pi}{12}\right)$$

$$=2\pi$$