Let, $I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx \dots \dots (1)$

Using formula,

$\int_{a}^{b} f(x) dx = \int_{a}^{b} f(a + b - x) dx$

$I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{\left[\cos\left(\frac{\pi}{2} - x\right)\right]^{2023}}{\left[\sin\left(\frac{\pi}{2} - x\right)\right]^{2023} + \left[\cos\left(\frac{\pi}{2} - x\right)\right]^{2023}} dx$

$= \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\sin x)^{2023}}{(\cos x)^{2023} + (\sin x)^{2023}} dx \dots \dots (2)$

Adding equation (1) and (2), we get

$2I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} \frac{(\sin x)^{2023} + (\cos x)^{2023}}{(\sin x)^{2023} + (\cos x)^{2023}} dx$

$\Rightarrow 2I = \frac{8}{\pi} \int_{0}^{\frac{\pi}{2}} dx$

$\Rightarrow 2I = \frac{8}{\pi} [x]_{0}^{\frac{\pi}{2}}$

$\Rightarrow 2I = \frac{8}{\pi} \times \frac{\pi}{2}$

$\Rightarrow 2I = 4$

$\Rightarrow I = 2$

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