Question Number 63666 by mathmax by abdo last updated on 07/Jul/19
$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{2}{sinx}\:+{cosx}} \\ $$
Commented by MJS last updated on 07/Jul/19
$$=\mathrm{0} \\ $$$$\mathrm{because}\:\mathrm{we}'\mathrm{re}\:\mathrm{integrating}\:\mathrm{over}\:\mathrm{a}\:\mathrm{full}\:\mathrm{period} \\ $$
Commented by mathmax by abdo last updated on 07/Jul/19
$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\mathrm{2}{sinx}\:+{cosx}}\:\:\:\:{changement}\:{e}^{{ix}} ={z}\:\:{give} \\ $$$${I}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{iz}\left(\:\mathrm{2}\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{1}\:+\frac{{iz}^{\mathrm{2}} +{i}}{\mathrm{2}}}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}\:+{iz}^{\mathrm{2}} \:+{i}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{\left(\mathrm{2}+{i}\right){z}^{\mathrm{2}} \:−\mathrm{2}+{i}}\:=\frac{\mathrm{2}}{\mathrm{2}+{i}}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}} \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}}\:\:{poles}\:{of}\:{W}? \\ $$$$\mid\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}\mid=\frac{\sqrt{\mathrm{5}}}{\:\sqrt{\mathrm{5}}}\:=\mathrm{1}\:{and}\:\:\:\:\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}\:=\frac{\sqrt{\mathrm{5}}{e}^{{iarctan}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\:\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:={e}^{−\mathrm{2}{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow \\ $$$$\sqrt{\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}}=\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\left({z}+{e}^{−{i}\:{actan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\:+{Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right\}\right. \\ $$$${Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)={lim}_{{z}\rightarrow−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} } \:\:\:\left({z}+{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right){W}\left({z}\right) \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2}{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:=−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\right)\right)} \\ $$$$=−\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$${Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\:=\frac{\mathrm{1}}{\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\right)\right)} =\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{−\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} +\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \right\}\:=\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\mathrm{0} \\ $$$$ \\ $$
Commented by MJS last updated on 07/Jul/19
$$\mathrm{2sin}\:{x}\:+\mathrm{cos}\:{x}\:=\sqrt{\mathrm{5}}\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{dt}}{\mathrm{sin}\:{t}}= \\ $$$$=−\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{t}}+\frac{\mathrm{1}}{\mathrm{tan}\:{t}}\right) \\ $$$$… \\ $$