Question:

Step 1: Define the given integral and identify the type of integral. Let the given expression be $E$. We need to evaluate the definite integral $I = \displaystyle\int_0^{\pi/2}\dfrac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\,dx$. The expression to be calculated is $\dfrac{8}{\pi}I$. This integral is of the form $\int_0^a \frac{f(x)}{f(x)+f(a-x)} dx$. Step 2: Apply the property of definite integrals. We use the property $\displaystyle\int_0^a f(x) dx = \displaystyle\int_0^a f(a-x) dx$. In this case, $a = \pi/2$. So, we replace $x$ with $(\pi/2 - x)$ in the integrand. Recall that $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$. Applying this property to $I$: $$I = \int_0^{\pi/2}\dfrac{(\cos(\pi/2 - x))^{2023}}{(\sin(\pi/2 - x))^{2023}+(\cos(\pi/2 - x))^{2023}}\,dx$$ $$I = \int_0^{\pi/2}\dfrac{(\sin x)^{2023}}{(\cos x)^{2023}+(\sin x)^{2023}}\,dx$$ Step 3: Add the original integral and the modified integral. Let the original integral be $I_1$ and the transformed integral be $I_2$. $$I_1 = \int_0^{\pi/2}\dfrac{(\cos x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\,dx$$ $$I_2 = \int_0^{\pi/2}\dfrac{(\sin x)^{2023}}{(\cos x)^{2023}+(\sin x)^{2023}}\,dx$$ Adding $I_1$ and $I_2$: $$2I = I_1 + I_2 = \int_0^{\pi/2}\dfrac{(\cos x)^{2023} + (\sin x)^{2023}}{(\sin x)^{2023}+(\cos x)^{2023}}\,dx$$ $$2I = \int_0^{\pi/2} 1\,dx$$ Step 4: Evaluate the simplified integral. Now, we evaluate the integral $2I$: $$2I = [x]_0^{\pi/2}$$ $$2I = \frac{\pi}{2} - 0$$ $$2I = \frac{\pi}{2}$$ Solving for $I$: $$I = \frac{\pi}{4}$$ Step 5: Calculate the final value of the given expression. The problem asks for the value of $\dfrac{8}{\pi}I$. Substitute the value of $I$ we found: $$E = \frac{8}{\pi} \cdot \left(\frac{\pi}{4}\right)$$ $$E = \frac{8}{4}$$ $$E = 2$$ The final answer is $\boxed{2}$.