Question Number 35687 by prof Abdo imad last updated on 22/May/18
$${calculate}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{1}−{a}\:{cosx}}\:\:{a}\:{from}\:{R}\:. \\ $$$$\left.\mathrm{2}\right)\:{application}\:\:{calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{cosx}} \\ $$
Commented by prof Abdo imad last updated on 22/May/18
$${vhangement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{a}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:−{a}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{a}\:\:+\left(\mathrm{1}+{a}\right){t}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}{t}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}>\mathrm{0}\:\:{and}\:{a}\neq\mathrm{1}\:\:\Rightarrow\:\frac{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}−{a}\right)}{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }>\mathrm{0}\:\Rightarrow \\ $$$$\:\mathrm{1}−{a}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\:\mid{a}\mid<\mathrm{1}\:\:\:{we}\:{get} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+\left(\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{t}\right)^{\mathrm{2}} } \\ $$$$=_{{u}=\:\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{t}} \:\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{1}−{a}}}{\:\sqrt{\mathrm{1}+{a}}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{1}−{a}}\sqrt{\mathrm{1}+{a}}}\:.\frac{\pi}{\mathrm{2}}\:=\:\:\frac{\pi}{\:\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$${case}\mathrm{2}\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}<\mathrm{0}\:\:\Rightarrow\:\frac{\mathrm{1}+{a}}{{a}−\mathrm{1}}>\mathrm{0}\:{and} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}\:\:−\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}{t}^{\mathrm{2}} } \\ $$$$=_{\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}\:{t}\:={u}} \:\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}\:−{u}^{\mathrm{2}} }\:\frac{\sqrt{{a}−\mathrm{1}}}{\:\sqrt{{a}+\mathrm{1}}}\:{du} \\ $$$$=\:−\frac{\mathrm{2}}{\:\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\int_{\mathrm{0}} ^{\infty} \left\{\:\frac{\mathrm{1}}{{u}−\mathrm{1}}\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right\}{du} \\ $$$$=\:\frac{\mathrm{1}}{\:\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\left[{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0}\:. \\ $$
Commented by prof Abdo imad last updated on 22/May/18
$$\left.\mathrm{2}\right){we}\:{have}\:{a}\:=\mathrm{2}\:{and}\:\mid{a}\mid>\mathrm{1}\:\Rightarrow\: \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{cosx}}\:=\mathrm{0} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18
$$\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{\mathrm{1}−{acosx}} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{\frac{\mathrm{1}}{{a}}−{cosx}} \\ $$$${let}\:{k}=\mathrm{1}/{a} \\ $$$$={k}\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{{k}−\frac{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}} \\ $$$$={k}\int_{\mathrm{0}} ^{\Pi} \frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}{{k}+{ktan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$${t}={tan}\frac{{x}}{\mathrm{2}}\:{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$={k}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{dt}}{\left({k}−\mathrm{1}\right)+\left({k}+\mathrm{1}\right){t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}{k}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({k}+\mathrm{1}\right)\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}×\frac{\mathrm{1}}{\:\sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}}}×\mid{tan}^{−\mathrm{1}} \left(\frac{{t}}{\:\sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}}}\:\:\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}×\frac{\sqrt{{k}+\mathrm{1}}\:}{\:\sqrt{{k}−\mathrm{1}}}×\left(\frac{\Pi}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{{k}}}×\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}}}}{\:\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{k}}\:}}×\left(\frac{\Pi}{\mathrm{2}}\right)\:\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{a}}×\frac{\sqrt{\mathrm{1}+{a}}}{\:\sqrt{\mathrm{1}−{a}}}×\left(\frac{\Pi}{\mathrm{2}}\right)\: \\ $$$$ \\ $$