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Question Number 152420 by tabata last updated on 28/Aug/21

∫_0 ^( (π/2)) (e^(sinx) /(e^(cosx) +e^(sinx) ))dx ?

$$\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \:\frac{{e}^{{sinx}} }{{e}^{{cosx}} +{e}^{{sinx}} }{dx}\:\:\:? \\ $$

Answered by mindispower last updated on 28/Aug/21

f(a+b−x)+f(x)=c ∫_a ^b f(x)dx=(1/2)c(b−a)

$${f}\left({a}+{b}−{x}\right)+{f}\left({x}\right)={c} \\ $$$$\int_{{a}} ^{{b}} {f}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{2}}{c}\left({b}−{a}\right) \\ $$

Answered by Lordose last updated on 28/Aug/21

Ω = ∫_0 ^(𝛑/2) (e^(sin(x)) /(e^(cos(x)) +e^(sin(x)) ))dx (1) Ω =^(king′s rule) ∫_0 ^(𝛑/2) (e^(cos(x)) /(e^(cos(x)) +e^(sin(x)) ))dx (2) (((1) + (2))/2) :: Ω = (1/2)∫_0 ^(𝛑/2) 1dx = (𝛑/4)

$$\Omega\:=\:\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} }{\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} +\mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} }\mathrm{dx}\:\:\:\:\left(\mathrm{1}\right) \\ $$$$\Omega\:\overset{\boldsymbol{\mathrm{king}}'\boldsymbol{\mathrm{s}}\:\boldsymbol{\mathrm{rule}}} {=}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \frac{\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} }{\mathrm{e}^{\mathrm{cos}\left(\mathrm{x}\right)} +\mathrm{e}^{\mathrm{sin}\left(\mathrm{x}\right)} }\mathrm{dx}\:\:\:\left(\mathrm{2}\right) \\ $$$$\frac{\left(\mathrm{1}\right)\:+\:\left(\mathrm{2}\right)}{\mathrm{2}}\::: \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\boldsymbol{\pi}}{\mathrm{2}}} \mathrm{1dx}\:=\:\frac{\boldsymbol{\pi}}{\mathrm{4}} \\ $$

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