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Question Number 116627 by mnjuly1970 last updated on 05/Oct/20

... nice calculus... please evaluate :: Φ = (((∫_0 ^( (π/2)) ((e^x +e^(−x) )/(sin(x)+cos(x)))dx)^2 )/((∫_0 ^( (π/2)) (e^x /(sin(x)+cos(x)))dx)(∫_0 ^( (π/2)) (e^(−x) /(sin(x)+cos(x)))dx))) =??? m.n.1970

$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:...\:\:{nice}\:\:\:{calculus}... \\ $$$$ \\ $$$$\:\:\:\:\:\:{please}\:\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Phi\:=\:\frac{\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} +{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)^{\mathrm{2}} }{\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)}\:=???\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$ \\ $$

Answered by mindispower last updated on 05/Oct/20

∫_0 ^(π/2) (e^x /(sin(x)+cos(x)))dx=J =∫_0 ^(π/2) (e^((π/2)−x) /(cos(x)+sin(x)))dx =e^(π/2) ∫_0 ^(π/2) (e^(−x) /(sin(x)+cos(x)))=e^(π/2) I Φ=(((J+I)^2 )/(I.J))=(((Ie^(π/2) +I)^2 )/(I.e^(π/2) I))=(((e^(π/2) +1)^2 )/e^(π/2) ) =e^(π/2) +e^(−(π/2)) +2=2(ch((π/2))+1)

$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}={J} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{\frac{\pi}{\mathrm{2}}−{x}} }{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx} \\ $$$$={e}^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}={e}^{\frac{\pi}{\mathrm{2}}} {I} \\ $$$$\Phi=\frac{\left({J}+{I}\right)^{\mathrm{2}} }{{I}.{J}}=\frac{\left({Ie}^{\frac{\pi}{\mathrm{2}}} +{I}\right)^{\mathrm{2}} }{{I}.{e}^{\frac{\pi}{\mathrm{2}}} {I}}=\frac{\left({e}^{\frac{\pi}{\mathrm{2}}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{\frac{\pi}{\mathrm{2}}} } \\ $$$$={e}^{\frac{\pi}{\mathrm{2}}} +{e}^{−\frac{\pi}{\mathrm{2}}} +\mathrm{2}=\mathrm{2}\left({ch}\left(\frac{\pi}{\mathrm{2}}\right)+\mathrm{1}\right) \\ $$

Commented by mnjuly1970 last updated on 05/Oct/20

vood for you

$${vood}\:{for}\:{you}\:\: \\ $$

Answered by Dwaipayan Shikari last updated on 05/Oct/20

I=∫_0 ^(π/2) ((e^x +e^(−x) )/(sinx+cosx))dx I_1 =∫_0 ^(π/2) (e^x /(sinx+cosx))dx=∫_0 ^(π/2) (e^((π/2)−x) /(sinx+cosx))dx=e^(π/2) ∫_0 ^(π/2) (e^(−x) /(sinx+cosx))dx I_2 =∫_0 ^(π/2) (e^(−x) /(sinx+cosx))dx I_2 =e^(π/2) I_1 I_1 +I_2 =I (By observation) Φ=(((I)^2 )/(I_1 I_2 ))=(((I_1 +I_2 )^2 )/(I_1 ^2 e^(π/2) ))=(((e^(π/2) +1)^2 )/e^(π/2) )=i^i .((i)^(1/i) +1)^2

$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} +{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sinx}+{cosx}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{\frac{\pi}{\mathrm{2}}−{x}} }{{sinx}+{cosx}}{dx}={e}^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{2}} ={e}^{\frac{\pi}{\mathrm{2}}} {I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{2}} ={I}\:\:\left({By}\:{observation}\right) \\ $$$$\Phi=\frac{\left({I}\right)^{\mathrm{2}} }{{I}_{\mathrm{1}} {I}_{\mathrm{2}} }=\frac{\left({I}_{\mathrm{1}} +{I}_{\mathrm{2}} \right)^{\mathrm{2}} }{{I}_{\mathrm{1}} ^{\mathrm{2}} {e}^{\frac{\pi}{\mathrm{2}}} }=\frac{\left({e}^{\frac{\pi}{\mathrm{2}}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{\frac{\pi}{\mathrm{2}}} }={i}^{{i}} .\left(\sqrt[{{i}}]{{i}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$

Commented by Dwaipayan Shikari last updated on 05/Oct/20

i^i =e^(−(π/2)) and (i)^(1/i) =i^(1/i) =i^(−i) =e^(π/2)

$$\mathrm{i}^{\mathrm{i}} =\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:{and}\:\sqrt[{{i}}]{{i}}={i}^{\frac{\mathrm{1}}{{i}}} ={i}^{−{i}} ={e}^{\frac{\pi}{\mathrm{2}}} \\ $$

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