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Question Number 163368 by nurtani last updated on 06/Jan/22

Answered by Ar Brandon last updated on 06/Jan/22

I=∫_0 ^(π/2) (((4/π)sin^(2007) x)/(sin^(2007) x+cos^(2007) x))dx...(1) I=∫_0 ^(π/2) (((4/π)cos^(2007) x)/(sin^(2007) x+cos^(2007) x))dx...(2) (1)+(2)⇒ 2I=(4/π)∫_0 ^(π/2) ((sin^(2007) x+cos^(2007) x)/(sin^(2007) x+cos^(2007) x))dx=(4/π)∫_0 ^(π/2) dx I=(2/π)∫_0 ^(π/2) dx=(2/π)∙(π/2)=1

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\mathrm{4}}{\pi}\mathrm{sin}^{\mathrm{2007}} {x}}{\mathrm{sin}^{\mathrm{2007}} {x}+\mathrm{cos}^{\mathrm{2007}} {x}}{dx}...\left(\mathrm{1}\right) \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\frac{\mathrm{4}}{\pi}\mathrm{cos}^{\mathrm{2007}} {x}}{\mathrm{sin}^{\mathrm{2007}} {x}+\mathrm{cos}^{\mathrm{2007}} {x}}{dx}...\left(\mathrm{2}\right) \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\Rightarrow \\ $$$$\mathrm{2}{I}=\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{sin}^{\mathrm{2007}} {x}+\mathrm{cos}^{\mathrm{2007}} {x}}{\mathrm{sin}^{\mathrm{2007}} {x}+\mathrm{cos}^{\mathrm{2007}} {x}}{dx}=\frac{\mathrm{4}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx} \\ $$$$\:\:{I}=\frac{\mathrm{2}}{\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {dx}=\frac{\mathrm{2}}{\pi}\centerdot\frac{\pi}{\mathrm{2}}=\mathrm{1} \\ $$

Commented by nurtani last updated on 06/Jan/22

thanks mr brandon

Commented by amin96 last updated on 06/Jan/22

great sir Brandon. Integral geniusss))

$$\left.{g}\left.{reat}\:{sir}\:{Brandon}.\:{Integral}\:{geniusss}\right)\right) \\ $$

Commented by Ar Brandon last updated on 06/Jan/22

My pleasure, friends!

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