Medium INTEGER +4 / -1 PYQ · JEE Mains 2023

The value of $${8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $$ is ___________

Answer (integer) 2

Solution

<p>Let, $$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $$ ..... (1)</p> <p>Using formula,</p> <p>$\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} }$</p> <p>$$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}} \over {{{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{2023}} + {{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}}}dx} $$</p> <p>$$ = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}}} \over {{{(\cos x)}^{2023}} + {{(\sin x)}^{2023}}}}dx} $$ ..... (2)</p> <p>Adding equation (1) and (2), we get</p> <p>$$2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{203}}}}dx} $$</p> <p>$\Rightarrow 2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {dx}$</p> <p>$\Rightarrow 2I = {8 \over \pi }\left[ x \right]_0^{{\pi \over 2}}$</p> <p>$\Rightarrow 2I = {8 \over \pi } \times {\pi \over 2}$</p> <p>$\Rightarrow 2I = 4$</p> <p>$\Rightarrow I = 2$</p>

About this question

Subject: Mathematics · Chapter: Definite Integration · Topic: Properties of Definite Integrals

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