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Question Number 29856 by abdo imad last updated on 13/Feb/18

find ∫_0 ^(2π) ((cos(nθ))/(2+3cosθ))dθ . n from N.

$${find}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}+\mathrm{3}{cos}\theta}{d}\theta\:.\:\:{n}\:{from}\:{N}. \\ $$

Commented by prof Abdo imad last updated on 18/Feb/18

let put I= ∫_0 ^(2π) ((cos(nθ))/(2+3cosθ))dθ I = Re( ∫_0 ^(2π) (e^(inθ) /(2+cosθ))dθ)=Re(w_n ) ch. e^(iθ) =z give w_n = ∫_(∣z∣=1) (z^n /(2 +((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) ((2z^n )/(iz(4 +z+z^(−1) )))dz =∫_(∣z∣=1) ((−2iz^n )/(4z +z^2 +1))dz =∫_(∣z∣=1) ((−2i z^n )/(z^2 +4z +1))dz let introduce the complex function ϕ(z)= ((−2iz^n )/(z^2 +4z +1)) poles of ϕ? Δ^′ = 2^2 −1 =3 ⇒z_1 =−2+(√3) and z_2 =−2−(√3) ∣z_1 ∣ −1=2−(√3) −1=1−(√3)<1 and ∣z_2 ∣ −1=2+(√3)−1=1+(√3)>1(to eliminate from residus) so ∫_(∣z∣=1) ϕ(z)dz=2iπ Res(ϕ,z_1 ) but we have ϕ(z)= ((−2i z^n )/((z−z_1 )(z−z_2 ))) ⇒ Res(ϕ,z_1 )= ((−2i z_1 ^n )/(z_1 −z_2 )) = ((−2i(−2+(√3) )^n )/(2(√3))) = ((−i(−2+(√3))^n )/(√3)) ∫_(∣z∣=1) ϕ(z)dz= ((2π)/(√3)) (−2+(√3))^n Re( ∫.....)= ((2π)/(√3))(−2+(√3))^n ⇒ I= ((2π)/(√3))(−2+(√3))^n .

$${let}\:{put}\:{I}=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}\left({n}\theta\right)}{\mathrm{2}+\mathrm{3}{cos}\theta}{d}\theta \\ $$$${I}\:=\:{Re}\left(\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{e}^{{in}\theta} }{\mathrm{2}+{cos}\theta}{d}\theta\right)={Re}\left({w}_{{n}} \right)\:{ch}.\:{e}^{{i}\theta} \:={z}\:{give} \\ $$$${w}_{{n}} =\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{z}^{{n}} }{\mathrm{2}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{z}^{{n}} }{{iz}\left(\mathrm{4}\:+{z}+{z}^{−\mathrm{1}} \right)}{dz} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{2}{iz}^{{n}} }{\mathrm{4}{z}\:+{z}^{\mathrm{2}} \:+\mathrm{1}}{dz}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{2}{i}\:{z}^{{n}} }{{z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}}{dz}\:{let} \\ $$$${introduce}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)=\:\:\frac{−\mathrm{2}{iz}^{{n}} }{{z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}}\:{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} =\:\mathrm{2}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{3}\:\Rightarrow{z}_{\mathrm{1}} =−\mathrm{2}+\sqrt{\mathrm{3}}\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{3}} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}=\mathrm{2}−\sqrt{\mathrm{3}}\:−\mathrm{1}=\mathrm{1}−\sqrt{\mathrm{3}}<\mathrm{1}\:{and} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}=\mathrm{2}+\sqrt{\mathrm{3}}−\mathrm{1}=\mathrm{1}+\sqrt{\mathrm{3}}>\mathrm{1}\left({to}\:{eliminate}\:{from}\right. \\ $$$$\left.{residus}\right)\:{so} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:{but}\:{we}\:{have} \\ $$$$\varphi\left({z}\right)=\:\frac{−\mathrm{2}{i}\:{z}^{{n}} }{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\:\frac{−\mathrm{2}{i}\:{z}_{\mathrm{1}} ^{{n}} }{{z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} }\:=\:\:\frac{−\mathrm{2}{i}\left(−\mathrm{2}+\sqrt{\mathrm{3}}\:\right)^{{n}} }{\mathrm{2}\sqrt{\mathrm{3}}} \\ $$$$=\:\:\:\frac{−{i}\left(−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} }{\sqrt{\mathrm{3}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:\left(−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} \\ $$$${Re}\left(\:\int.....\right)=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\left(−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} \Rightarrow \\ $$$${I}=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\left(−\mathrm{2}+\sqrt{\mathrm{3}}\right)^{{n}} \:\:. \\ $$

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