Integration Questions

Question Number 84810 by M±th+et£s last updated on 16/Mar/20

$$\int_{\mathrm{0}} ^{\pi} {ln}\left(\frac{\mathrm{1}+{b}\:{cos}\left({x}\right)}{\mathrm{1}+{a}\:{sin}\left({x}\right)}\right)\:{dx} \\$$ $$−\mathrm{1}<{a}<{b}<\mathrm{1} \\$$

Commented bymathmax by abdo last updated on 16/Mar/20

$$\int_{\mathrm{0}} ^{\pi} {ln}\left(\frac{\mathrm{1}+{bcosx}}{\mathrm{1}+{asinx}}\right){dx}\:={g}\left({b}\right)−{f}\left({a}\right)\:{with}\:{g}\left({b}\right)=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{bcosx}\right){dx} \\$$ $${and}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{asinx}\right){dx}\:\:{we}\:{have}\:{g}^{'} \left({b}\right)=\int_{\mathrm{0}} ^{\pi} \:\frac{{cosx}}{\mathrm{1}+{bcosx}}{dx} \\$$ $$=\frac{\mathrm{1}}{{b}}\int_{\mathrm{0}} ^{\pi} \:\frac{{bcosx}+\mathrm{1}−\mathrm{1}}{\mathrm{1}+{bcosx}}{dx}\:=\frac{\pi}{{b}}−\frac{\mathrm{1}}{{b}}\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{bcosx}} \\$$ $$\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{1}+{bcosx}}\:=_{{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}} \:\:\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{1}}{\mathrm{1}+{b}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:+{b}−{bt}^{\mathrm{2}} } \\$$ $$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}−{b}\right){t}^{\mathrm{2}} \:+\mathrm{1}+{b}}\:=\frac{\mathrm{2}}{\mathrm{1}−{b}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dt}}{{t}^{\mathrm{2}} \:+\frac{\mathrm{1}+{b}}{\mathrm{1}−{b}}} \\$$ $$=_{{t}=\sqrt{\frac{\mathrm{1}+{b}}{\mathrm{1}−{b}}}{u}} \:\:\frac{\mathrm{2}}{\mathrm{1}−{b}}×\frac{\mathrm{1}−{b}}{\mathrm{1}+{b}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\sqrt{\frac{\mathrm{1}+{b}}{\mathrm{1}−{b}}}{du} \\$$ $$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }}×\frac{\pi}{\mathrm{2}}\:=\frac{\pi}{\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }}\:\Rightarrow{g}^{'} \left({b}\right)=\frac{\pi}{{b}}−\frac{\pi}{{b}\sqrt{\mathrm{1}−{b}^{\mathrm{2}} }}\:\Rightarrow \\$$ $${g}\left({b}\right)=\pi{ln}\mid{b}\mid−\pi\:\int_{{b}} ^{\mathrm{1}} \:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+{K} \\$$ $${g}\left(\mathrm{1}\right)\:={K} \\$$ $${K}=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx}\:\Rightarrow{g}\left({b}\right)\:=\pi{ln}\mid{b}\mid−\pi\:\int_{{b}} ^{\mathrm{1}} \:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:+\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx} \\$$ $$\int_{{b}} ^{\mathrm{1}} \:\frac{{dx}}{{x}\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\:=_{{x}={sin}\alpha} \:\:\:\int_{{arcsinb}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{cos}\alpha\:{d}\alpha}{{sin}\alpha.{cos}\alpha}\:=\int_{{arcsinb}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{d}\alpha}{{sin}\alpha} \\$$ $$=_{{tan}\left(\frac{\alpha}{\mathrm{2}}\right)={z}} \:\:\:\int_{{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)} ^{\mathrm{1}} \:\:\frac{\mathrm{2}{dz}}{\left(\mathrm{1}+{z}^{\mathrm{2}} \right)\frac{\mathrm{2}{z}}{\mathrm{1}+{z}^{\mathrm{2}} }}\:=\int_{{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)} ^{\mathrm{1}} \:\:\:\frac{{dz}}{{z}} \\$$ $$\left.=\left[{ln}\mid{z}\mid\right]_{{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)} ^{\mathrm{1}} \:=−{ln}\mid{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)\right)\:\Rightarrow \\$$ $${g}\left({b}\right)=\pi{ln}\mid{b}\mid+\pi{ln}\mid{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)\mid+\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx} \\$$ $$\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{1}+{cosx}\right){dx}\:=\int_{\mathrm{0}} ^{\pi} {ln}\left(\mathrm{2}{cos}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)\right){dx} \\$$ $$=\pi{ln}\left(\mathrm{2}\right)+\mathrm{2}\int_{\mathrm{0}} ^{\pi} \:{ln}\left({cos}\left(\frac{{x}}{\mathrm{2}}\right)\right){dx}\:\:\:\:\left(\frac{{x}}{\mathrm{2}}={u}\right) \\$$ $$=\pi{ln}\left(\mathrm{2}\right)+\mathrm{2}\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({cosu}\right)\left(\mathrm{2}{du}\right) \\$$ $$=\pi{ln}\left(\mathrm{2}\right)+\mathrm{4}\left(−\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right)\right)\:=−\pi{ln}\left(\mathrm{2}\right)\:\Rightarrow \\$$ $${g}\left({b}\right)=\pi\left({ln}\mid{b}\mid−{ln}\left(\mathrm{2}\right)\right)+\pi{ln}\mid{tan}\left(\frac{{arcsinb}}{\mathrm{2}}\right)\mid...{be}\:{continued} \\$$