QJEEM23Ma10b1: JEE Main 2023 Mathematics - Class 12 JEEM23Ma10b1 Solution

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

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

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

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

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

$\Rightarrow 2I = 4$

$\Rightarrow I = 2$