Let, $$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $$ ..... (1)

Using formula,

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

$$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}} \over {{{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{2023}} + {{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}}}dx} $$

$$ = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}}} \over {{{(\cos x)}^{2023}} + {{(\sin x)}^{2023}}}}dx} $$ ..... (2)

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

$$2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{203}}}}dx} $$

$$ \Rightarrow 2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {dx} $$

$$ \Rightarrow 2I = {8 \over \pi }\left[ x \right]_0^{{\pi \over 2}}$$

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

$$ \Rightarrow 2I = 4$$

$$ \Rightarrow I = 2$$