Question Number 116627 by mnjuly1970 last updated on 05/Oct/20
$$\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:…\:\:{nice}\:\:\:{calculus}… \\ $$$$ \\ $$$$\:\:\:\:\:\:{please}\:\:\:{evaluate}\::: \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\Phi\:=\:\frac{\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} +{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)^{\mathrm{2}} }{\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)\left(\int_{\mathrm{0}} ^{\:\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}\right)}\:=???\:\:\:\: \\ $$$$ \\ $$$$\:\:\:\:{m}.{n}.\mathrm{1970} \\ $$$$ \\ $$
Answered by mindispower last updated on 05/Oct/20
$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}{dx}={J} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{\frac{\pi}{\mathrm{2}}−{x}} }{{cos}\left({x}\right)+{sin}\left({x}\right)}{dx} \\ $$$$={e}^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sin}\left({x}\right)+{cos}\left({x}\right)}={e}^{\frac{\pi}{\mathrm{2}}} {I} \\ $$$$\Phi=\frac{\left({J}+{I}\right)^{\mathrm{2}} }{{I}.{J}}=\frac{\left({Ie}^{\frac{\pi}{\mathrm{2}}} +{I}\right)^{\mathrm{2}} }{{I}.{e}^{\frac{\pi}{\mathrm{2}}} {I}}=\frac{\left({e}^{\frac{\pi}{\mathrm{2}}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{\frac{\pi}{\mathrm{2}}} } \\ $$$$={e}^{\frac{\pi}{\mathrm{2}}} +{e}^{−\frac{\pi}{\mathrm{2}}} +\mathrm{2}=\mathrm{2}\left({ch}\left(\frac{\pi}{\mathrm{2}}\right)+\mathrm{1}\right) \\ $$
Commented by mnjuly1970 last updated on 05/Oct/20
$${vood}\:{for}\:{you}\:\: \\ $$
Answered by Dwaipayan Shikari last updated on 05/Oct/20
$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} +{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{{x}} }{{sinx}+{cosx}}{dx}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{\frac{\pi}{\mathrm{2}}−{x}} }{{sinx}+{cosx}}{dx}={e}^{\frac{\pi}{\mathrm{2}}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{2}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{e}^{−{x}} }{{sinx}+{cosx}}{dx} \\ $$$${I}_{\mathrm{2}} ={e}^{\frac{\pi}{\mathrm{2}}} {I}_{\mathrm{1}} \\ $$$${I}_{\mathrm{1}} +{I}_{\mathrm{2}} ={I}\:\:\left({By}\:{observation}\right) \\ $$$$\Phi=\frac{\left({I}\right)^{\mathrm{2}} }{{I}_{\mathrm{1}} {I}_{\mathrm{2}} }=\frac{\left({I}_{\mathrm{1}} +{I}_{\mathrm{2}} \right)^{\mathrm{2}} }{{I}_{\mathrm{1}} ^{\mathrm{2}} {e}^{\frac{\pi}{\mathrm{2}}} }=\frac{\left({e}^{\frac{\pi}{\mathrm{2}}} +\mathrm{1}\right)^{\mathrm{2}} }{{e}^{\frac{\pi}{\mathrm{2}}} }={i}^{{i}} .\left(\sqrt[{{i}}]{{i}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$
Commented by Dwaipayan Shikari last updated on 05/Oct/20
$$\mathrm{i}^{\mathrm{i}} =\mathrm{e}^{−\frac{\pi}{\mathrm{2}}} \:\:\:{and}\:\sqrt[{{i}}]{{i}}={i}^{\frac{\mathrm{1}}{{i}}} ={i}^{−{i}} ={e}^{\frac{\pi}{\mathrm{2}}} \\ $$