Question Number 27684 by abdo imad last updated on 12/Jan/18
$$\left.\mathrm{1}\right)\:{prove}\:{the}\:{existence}\:{of}\:{the}\:{integral} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}}{dx} \\ $$$$\left.\mathrm{2}\right){prove}\:{that}\:{I}=\:\int\int_{{D}} \:\:\frac{{siny}}{\mathrm{1}+{cosx}\:{cosy}}{dxdy}\:{with}\: \\ $$$${D}=\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:{I}. \\ $$
Commented by abdo imad last updated on 16/Jan/18
$$\left.\mathrm{1}\right)\:{the}\:{convergence}\:{of}\:{I}\:{is}\:{easy}\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}} \\ $$$$=\:{lim}_{{x}\rightarrow\frac{\pi}{\mathrm{2}}} \:\:\:\frac{−{sinx}}{−{sinx}\left(\mathrm{1}+{cosx}\right)}=\:\mathrm{1}\:\:\left({by}\:{hospital}\:{theorem}\:\:\right) \\ $$$${so}\:{the}\:{function}\:{is}\:{local}\:{integrable}\:{in}\:\left[\mathrm{0},\frac{\pi}{\mathrm{2}}\left[\right.\right. \\ $$$$\left.\mathrm{2}\right)\:{I}=\:\int\int_{\mathrm{0}\leqslant{x}\leqslant\:\frac{\pi}{\mathrm{2}}\:{and}\:\mathrm{0}\leqslant{y}\leqslant\:\frac{\pi}{\mathrm{2}}} \:\:\frac{{siny}}{\mathrm{1}+{cosx}\:{cosy}}\:{dx}\:{dy} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left(\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\mathrm{1}}{{cosx}}\left(\:\:−\left(\:\frac{\left.\mathrm{1}+{cosx}\:{cosy}\right)^{,} }{\mathrm{1}+{cosx}\:{cosy}}\right){dy}\right){dx}\right. \\ $$$$\left.{I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{\mathrm{1}}{{cosx}}\:\left[−{ln}/\mathrm{1}+{cosx}\:{cosy}/\right]_{{y}=\mathrm{0}} ^{{y}=\frac{\pi}{\mathrm{2}}} \right]{dx} \\ $$$${I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}\:}} \:\:\:\:\frac{{ln}\left(\mathrm{1}+{cosx}\right)}{{cosx}}{dx} \\ $$$$\left.\mathrm{3}\right)\:{I}=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cosy}\:{cosx}}\right){siny}\:{dy}\:\:{but} \\ $$$$\:{J}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{dx}}{\mathrm{1}+{cosy}\:{cosx}}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\frac{{dx}}{\mathrm{1}+\lambda\:{cosx}}\:\left(\lambda={cosy}\right){and}\:{the}\:{ch}. \\ $$$${tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:\:{give}\:\:{J}=\:\int_{\mathrm{0}^{} } ^{\mathrm{1}} \:\:\:\:\:\frac{\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{\mathrm{1}+\lambda\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+\lambda−\lambda{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}−\lambda\right){t}^{\mathrm{2}} \:+\mathrm{1}+\lambda}=\:\frac{\mathrm{2}}{\mathrm{1}−\lambda}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{{dt}}{{t}^{\mathrm{2}} +\left(\sqrt{\left.\frac{\mathrm{1}+\lambda}{\mathrm{1}−\lambda}\right)}\right.}\mathrm{2} \\ $$$$=\:\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:\int_{\mathrm{0}} ^{\sqrt{\frac{\mathrm{1}−\lambda}{\mathrm{1}+\lambda}}} \:\:\:\:\:\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\:\:\:\:\:\left({ch}.{t}=\sqrt{\frac{\mathrm{1}+\lambda}{\mathrm{1}−\lambda}}\:\:{u}\right) \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−\lambda^{\mathrm{2}} }}\:{artan}\left(\:\sqrt{\frac{\mathrm{1}−\lambda}{\left.\mathrm{1}+\lambda\right)}}\right)=\:\frac{\mathrm{2}}{{siny}}\:{artan}\left({tan}\left(\frac{{y}}{\left.\mathrm{2}\right)}\right)\right)=\:\frac{{y}}{{siny}} \\ $$$${I}=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{{y}}{{siny}}\:{sinydy}=\:\left[\frac{\mathrm{1}}{\mathrm{2}}\:{y}^{\mathrm{2}} \right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} =\:\frac{\mathrm{1}}{\mathrm{2}}\:\frac{\pi^{\mathrm{2}} }{\mathrm{4}}=\:\frac{\pi^{\mathrm{2}} }{\mathrm{8}}\:. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$