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Question Number 63667 by mathmax by abdo last updated on 07/Jul/19

1) calculate ∫_0 ^(2π) (dt/(cost +x sint)) wih x from R. 2) calculate ∫_0 ^(2π) ((sint)/((cost +xsint)^2 ))dt 3) find[the value of ∫_0 ^(2π) (dt/(cos(2t)+2sin(2t)))

$$\left.\mathrm{1}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{cost}\:+{x}\:{sint}}\:\:\:{wih}\:{x}\:{from}\:{R}. \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right)\:{find}\left[{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)+\mathrm{2}{sin}\left(\mathrm{2}{t}\right)}\right. \\ $$

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

1) changement z =e^(it) give ∫_0 ^(2π) (dt/(cost +xsint)) =∫_(∣z∣=1) (dz/(iz{((z+z^(−1) )/2)+x((z−z^(−1) )/(2i))})) =∫_(∣z∣=1) (dz/(((iz^2 +i)/2)+((xz^2 −x)/2))) =∫_(∣z∣=1) ((2dz)/((x+i)z^2 −x+i)) =(2/(x+i)) ∫_(∣z∣=1) (dz/(z^2 −((x−i)/(x+i)))) we have ((x−i)/(x+i)) =(((√(x^2 +1))e^(iarctan(−(1/x))) )/((√(x^2 +1))e^(iarctan((1/x))) )) =e^(−2i arctan((1/x))) ( we suppose x≠0) ⇒(√((x−i)/(x+i)))=e^(−iarctan((1/x))) let w(z) =(1/(z^2 −((x−i)/(x+i)))) ⇒W(z)=(1/((z−e^(−iarctan((1/x))) )(z+e^(−iarctan((1/x))) ))) residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ{Res(W,e^(−iarctan((1/x))) )+Res(W,−e^(−iarctan((1/x))) ) Res(W,e^(−iarctan((1/x))) )=(1/(2e^(−iarctan((1/x))) )) =(1/2) e^(iarctan((1/x))) Res(W,−e^(−iarctan((1/x))) ) =(1/(−2e^(−iarctan((1/x))) )) =−(1/2)e^(iarctan((1/x))) ⇒ the residus are opposites ⇒∫_(∣z∣=1) W(z)dz =0 ⇒ ∀x ≠0 ∫_0 ^(2π) (dt/(cost +isint)) =0 if x=0 we get ∫_0 ^(2π) (dt/(cost)) =∫_0 ^π (dt/(cost)) +∫_π ^(2π) (dt/(cost)) ∫_π ^(2π) (dt/(cost)) =_(t =π +u) ∫_0 ^π (du/(−cosu)) =−∫_0 ^π (du/(cosu)) ⇒ ∫_0 ^(2π) (dt/(cost)) =0

$$\left.\mathrm{1}\right)\:{changement}\:{z}\:={e}^{{it}} \:{give}\: \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}\:+{xsint}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{dz}}{{iz}\left\{\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+{x}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}\right\}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{dz}}{\frac{{iz}^{\mathrm{2}} +{i}}{\mathrm{2}}+\frac{{xz}^{\mathrm{2}} −{x}}{\mathrm{2}}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{\left({x}+{i}\right){z}^{\mathrm{2}} −{x}+{i}} \\ $$$$=\frac{\mathrm{2}}{{x}+{i}}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{dz}}{{z}^{\mathrm{2}} −\frac{{x}−{i}}{{x}+{i}}}\:\:\:\:\:\:{we}\:{have}\:\frac{{x}−{i}}{{x}+{i}}\:=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{e}^{{iarctan}\left(−\frac{\mathrm{1}}{{x}}\right)} }{\sqrt{{x}^{\mathrm{2}} \:+\mathrm{1}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} } \\ $$$$={e}^{−\mathrm{2}{i}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\:\:\:\left(\:\:{we}\:{suppose}\:{x}\neq\mathrm{0}\right)\:\Rightarrow\sqrt{\frac{{x}−{i}}{{x}+{i}}}={e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${let}\:{w}\left({z}\right)\:=\frac{\mathrm{1}}{{z}^{\mathrm{2}} −\frac{{x}−{i}}{{x}+{i}}}\:\Rightarrow{W}\left({z}\right)=\frac{\mathrm{1}}{\left({z}−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\left({z}+{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)} \\ $$$${residus}\:{theorem}\:{give}\: \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:{W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{{Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)+{Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\right. \\ $$$${Res}\left({W},{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)=\frac{\mathrm{1}}{\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} }\:=\frac{\mathrm{1}}{\mathrm{2}}\:{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \\ $$$${Res}\left({W},−{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \right)\:=\frac{\mathrm{1}}{−\mathrm{2}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} }\:=−\frac{\mathrm{1}}{\mathrm{2}}{e}^{{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$${the}\:{residus}\:{are}\:{opposites}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{0}\:\Rightarrow \\ $$$$\forall{x}\:\neq\mathrm{0}\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{{cost}\:+{isint}}\:=\mathrm{0} \\ $$$${if}\:{x}=\mathrm{0}\:{we}\:{get}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}}\:=\int_{\mathrm{0}} ^{\pi} \:\frac{{dt}}{{cost}}\:+\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}} \\ $$$$\int_{\pi} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}}\:=_{{t}\:=\pi\:+{u}} \:\:\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{du}}{−{cosu}}\:=−\int_{\mathrm{0}} ^{\pi} \:\frac{{du}}{{cosu}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cost}}\:=\mathrm{0} \\ $$

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

2) let f(x)=∫_0 ^(2π) (dt/(cost +xsint)) ⇒f^′ (x) =−∫_0 ^(2π) ((sint)/((cost +xsint)^2 ))dt =0 (because f(x)=0)⇒∫_0 ^(2π) ((sint)/((cost +xsint)^2 ))dt =0

$$\left.\mathrm{2}\right)\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{cost}\:+{xsint}}\:\Rightarrow{f}^{'} \left({x}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt}\:=\mathrm{0} \\ $$$$\left({because}\:{f}\left({x}\right)=\mathrm{0}\right)\Rightarrow\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sint}}{\left({cost}\:+{xsint}\right)^{\mathrm{2}} }{dt}\:=\mathrm{0} \\ $$

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

3)∫_0 ^(2π) (dt/(cos(2t)+2 sin(2t))) =_(2t =u) ∫_0 ^(4π) (du/(2(cosu +2sinu))) =(1/2)∫_0 ^(2π) (du/(cosu +2sinu )) +∫_(2π) ^(4π) (du/(cosu +2sinu)) =0 because 2π is period.

$$\left.\mathrm{3}\right)\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{{cos}\left(\mathrm{2}{t}\right)+\mathrm{2}\:{sin}\left(\mathrm{2}{t}\right)}\:=_{\mathrm{2}{t}\:={u}} \:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\frac{{du}}{\mathrm{2}\left({cosu}\:+\mathrm{2}{sinu}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{du}}{{cosu}\:+\mathrm{2}{sinu}\:}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\frac{{du}}{{cosu}\:+\mathrm{2}{sinu}}\:=\mathrm{0}\:{because} \\ $$$$\mathrm{2}\pi\:{is}\:{period}. \\ $$

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