Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

Question Number 63666 by mathmax by abdo last updated on 07/Jul/19

calculate ∫_0 ^(2π) (dx/(2sinx +cosx))

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{2}{sinx}\:+{cosx}} \\ $$

Commented by MJS last updated on 07/Jul/19

=0 because we′re integrating over a full period

$$=\mathrm{0} \\ $$$$\mathrm{because}\:\mathrm{we}'\mathrm{re}\:\mathrm{integrating}\:\mathrm{over}\:\mathrm{a}\:\mathrm{full}\:\mathrm{period} \\ $$

Commented by mathmax by abdo last updated on 07/Jul/19

let I =∫_0 ^(2π) (dx/(2sinx +cosx)) changement e^(ix) =z give I =∫_(∣z∣=1) (dz/(iz( 2 ((z−z^(−1) )/(2i))+((z+z^(−1) )/2)))) =∫_(∣z∣=1) (dz/(z^2 −1 +((iz^2 +i)/2))) = ∫_(∣z∣=1) ((2dz)/(2z^2 −2 +iz^2 +i)) =∫_(∣z∣=1) ((2dz)/((2+i)z^2 −2+i)) =(2/(2+i)) ∫_(∣z∣=1) (dz/(z^2 −((2−i)/(2+i)))) let W(z)=(1/(z^2 −((2−i)/(2+i)))) poles of W? ∣((2−i)/(2+i))∣=((√5)/(√5)) =1 and ((2−i)/(2+i)) =(((√5)e^(iarctan(−(1/2))) )/((√5)e^(iarctan((1/2))) )) =e^(−2i arctan((1/2))) ⇒ (√((2−i)/(2+i)))= e^(−i arctan((1/2))) ⇒W(z) =(1/((z−e^(−iarctan((1/2))) )(z+e^(−i actan((1/2))) ))) residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ {Res(W,−e^(−iarctan((1/2))) ) +Res(W,e^(−iarctan((1/2))) } Res(W,−e^(−iarctan((1/2))) )=lim_(z→−e^(−iarctan((1/2))) ) (z+e^(−i arctan((1/2))) )W(z) =(1/(−2e^(−i arctan((1/2))) )) =−(1/2) e^(i arctan((1/2))) =−(1/2) e^(i((π/2)−arctan(2))) =−(i/2) e^(−iarctan(2)) ⇒ Res(W,e^(−iarctan((1/2))) ) =(1/(2e^(−iarctan((1/2))) )) =(1/2) e^(i arctan((1/2))) =(1/2) e^(i((π/2)−arctan(2))) =(i/2) e^(−iarctan(2)) ⇒ ∫_(∣z∣=1) W(z)dz =2iπ{−(i/2) e^(−iarctan(2)) +(i/2) e^(−iarctan(2)) } =0 ⇒ I =0

$${let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\mathrm{2}{sinx}\:+{cosx}}\:\:\:\:{changement}\:{e}^{{ix}} ={z}\:\:{give} \\ $$$${I}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{iz}\left(\:\mathrm{2}\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\mathrm{1}\:+\frac{{iz}^{\mathrm{2}} +{i}}{\mathrm{2}}}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{\mathrm{2}{z}^{\mathrm{2}} −\mathrm{2}\:+{iz}^{\mathrm{2}} \:+{i}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{dz}}{\left(\mathrm{2}+{i}\right){z}^{\mathrm{2}} \:−\mathrm{2}+{i}}\:=\frac{\mathrm{2}}{\mathrm{2}+{i}}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}} \\ $$$${let}\:{W}\left({z}\right)=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}}\:\:{poles}\:{of}\:{W}? \\ $$$$\mid\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}\mid=\frac{\sqrt{\mathrm{5}}}{\sqrt{\mathrm{5}}}\:=\mathrm{1}\:{and}\:\:\:\:\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}\:=\frac{\sqrt{\mathrm{5}}{e}^{{iarctan}\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)} }{\sqrt{\mathrm{5}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:={e}^{−\mathrm{2}{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow \\ $$$$\sqrt{\frac{\mathrm{2}−{i}}{\mathrm{2}+{i}}}=\:{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:\Rightarrow{W}\left({z}\right)\:=\frac{\mathrm{1}}{\left({z}−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\left({z}+{e}^{−{i}\:{actan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\:+{Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right\}\right. \\ $$$${Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)={lim}_{{z}\rightarrow−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} } \:\:\:\left({z}+{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right){W}\left({z}\right) \\ $$$$=\frac{\mathrm{1}}{−\mathrm{2}{e}^{−{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:=−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \:=−\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\right)\right)} \\ $$$$=−\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$${Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \right)\:=\frac{\mathrm{1}}{\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\:{arctan}\left(\frac{\mathrm{1}}{\mathrm{2}}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{i}\left(\frac{\pi}{\mathrm{2}}−{arctan}\left(\mathrm{2}\right)\right)} =\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{−\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} +\frac{{i}}{\mathrm{2}}\:{e}^{−{iarctan}\left(\mathrm{2}\right)} \right\}\:=\mathrm{0}\:\Rightarrow \\ $$$${I}\:=\mathrm{0} \\ $$$$ \\ $$

Commented by MJS last updated on 07/Jul/19

2sin x +cos x =(√5)sin (x+arctan (1/2)) ((√5)/5)∫(dx/(sin (x+arctan (1/2))))=((√5)/5)∫(dt/(sin t))= =−ln ((1/(sin t))+(1/(tan t))) ...

$$\mathrm{2sin}\:{x}\:+\mathrm{cos}\:{x}\:=\sqrt{\mathrm{5}}\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{dx}}{\mathrm{sin}\:\left({x}+\mathrm{arctan}\:\frac{\mathrm{1}}{\mathrm{2}}\right)}=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\int\frac{{dt}}{\mathrm{sin}\:{t}}= \\ $$$$=−\mathrm{ln}\:\left(\frac{\mathrm{1}}{\mathrm{sin}\:{t}}+\frac{\mathrm{1}}{\mathrm{tan}\:{t}}\right) \\ $$$$... \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com