Question Number 63667 by mathmax by abdo last updated on 07/Jul/19
$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{cost}\:+{x}\:{sint}}\:\:\:{wih}\:{x}\:{from}\:{R}. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\left[{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)+\mathrm{2}{sin}\left(\mathrm{2}{t}\right)}\right. \\ $$
Commented by mathmax by abdo last updated on 07/Jul/19
$$\left.\mathrm{1}\right)\:{changement}\:{z}\:={e}^{{it}} \:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}\:+{xsint}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{iz}\left\{\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+{x}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}\right\}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{dz}}{\frac{{iz}^{\mathrm{2}} +{i}}{\mathrm{2}}+\frac{{xz}^{\mathrm{2}} −{x}}{\mathrm{2}}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{\left({x}+{i}\right){z}^{\mathrm{2}} −{x}+{i}} \\ $$$$=\frac{\mathrm{2}}{{x}+{i}}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\frac{{x}−{i}}{{x}+{i}}}\:\:\:\:\:\:{we}\:{have}\:\frac{{x}−{i}}{{x}+{i}}\:=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{e}^{{iarctan}\left(−\frac{\mathrm{1}}{{x}}\right)} }{\:\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} } \\ $$$$={e}^{−\mathrm{2}{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\:\:\:\left(\:\:{we}\:{suppose}\:{x}\neq\mathrm{0}\right)\:\Rightarrow\sqrt{\frac{{x}−{i}}{{x}+{i}}}={e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\frac{{x}−{i}}{{x}+{i}}}\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\left({z}+{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give}\: \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)+{Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\right. \\ $$$${Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)=\frac{\mathrm{1}}{\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\:=\frac{\mathrm{1}}{−\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} }\:=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$${the}\:{residus}\:{are}\:{opposites}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{0}\:\Rightarrow \\ $$$$\forall{x}\:\neq\mathrm{0}\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{cost}\:+{isint}}\:=\mathrm{0} \\ $$$${if}\:{x}=\mathrm{0}\:{we}\:{get}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dt}}{{cost}}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}} \\ $$$$\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}}\:=_{{t}\:=\pi\:+{u}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{−{cosu}}\:=−\int_{\mathrm{0}} ^{\pi} \:\frac{{du}}{{cosu}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}}\:=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 07/Jul/19
$$\left.\mathrm{2}\right)\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}\:+{xsint}}\:\Rightarrow{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt}\:=\mathrm{0} \\ $$$$\left({because}\:{f}\left({x}\right)=\mathrm{0}\right)\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt}\:=\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 07/Jul/19
$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)+\mathrm{2}\:{sin}\left(\mathrm{2}{t}\right)}\:=_{\mathrm{2}{t}\:={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{{du}}{\mathrm{2}\left({cosu}\:+\mathrm{2}{sinu}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{du}}{{cosu}\:+\mathrm{2}{sinu}\:}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\frac{{du}}{{cosu}\:+\mathrm{2}{sinu}}\:=\mathrm{0}\:{because} \\ $$$$\mathrm{2}\pi\:{is}\:{period}. \\ $$