1. Let $I=\int_0^{\pi/2} \frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\,dx.$

We need to find $\frac{8}{\pi}I.$

2. Use the standard substitution $x\mapsto \frac{\pi}{2}-x.$ Then $\sin\left(\frac{\pi}{2}-x\right)=\cos x,\qquad \cos\left(\frac{\pi}{2}-x\right)=\sin x.$ So, $I=\int_0^{\pi/2} \frac{(\sin x)^{2023}}{(\cos x)^{2023}+(\sin x)^{2023}}\,dx.$

3. Add the two expressions for $I$: $2I=\int_0^{\pi/2}\left[\frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}+\frac{(\sin x)^{2023}}{(\cos x)^{2023}+(\sin x)^{2023}}\right]dx.$

Inside the bracket, $\frac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}+\frac{(\sin x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}=1.$ Therefore, $2I=\int_0^{\pi/2}1\,dx=\frac\pi2.$ Hence, $I=\frac\pi4.$

4. Now compute the required value: $\frac{8}{\pi}I=\frac{8}{\pi}\cdot \frac\pi4=2.$

Therefore, the value of the given expression is $\boxed{2}.$

5. Comparison with stored correct answer:

• Derived answer: $2$
• Stored correct answer: $2$
• They match.