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Question Number 31513 by abdo imad last updated on 09/Mar/18

find ∫_0 ^(2π) (dx/(2 +cosx)) .

$${find}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\mathrm{2}\:+{cosx}}\:\:. \\ $$

Commented by abdo imad last updated on 10/Mar/18

the ch. e^(ix) =z give I=∫_(∣z∣=1) (1/(2+((z+z^(−1) )/2))) (dz/(iz))= ∫_(∣z∣=1) ((2dz)/(iz(4 +z +z^(−1) ))) = ∫_(∣z∣=1) ((−2i dz)/(4z +z^2 +1)) = ∫_(∣z∣=1) ((−2idz)/(z^2 +4z +1)) let consider ϕ(z)= ((−2i)/(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) <0 ⇒ ∣z_1 ∣<1 ∣z_2 ∣ −1=2+(√3) −1=1+(√3) >1 (to eliminate from residus) ∫_(∣z∣=1) ϕ(z)dz=2iπ Res(ϕ,z_1 ) but ϕ(z)= ((−2i)/((z−z_1 )(z−z_2 ))) ⇒ Res(ϕ,z_1 )= ((−2i)/(z_1 −z_2 ))= ((−2i)/(2(√3))) =((−i)/(√3)) ∫_(∣z∣=1) ϕ(z)dz=2iπ .((−i)/(√3))=((2π)/((√3) )) ⇒ I= ((2π)/(√3)) .

$${the}\:{ch}.\:{e}^{{ix}} ={z}\:{give} \\ $$$${I}=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{4}\:+{z}\:+{z}^{−\mathrm{1}} \right)} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{2}{i}\:{dz}}{\mathrm{4}{z}\:+{z}^{\mathrm{2}} \:+\mathrm{1}}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{2}{idz}}{{z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}}\:{let}\:{consider} \\ $$$$\varphi\left({z}\right)=\:\frac{−\mathrm{2}{i}}{{z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}}\:\:{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} =\mathrm{2}^{\mathrm{2}} \:−\mathrm{1}=\mathrm{3}\:\:\Rightarrow\:{z}_{\mathrm{1}} =−\mathrm{2}\:+\sqrt{\mathrm{3}}\:\:{and}\:{z}_{\mathrm{2}} =−\mathrm{2}−\sqrt{\mathrm{3}}\:\: \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}=\mathrm{2}−\sqrt{\mathrm{3}}\:−\mathrm{1}=\mathrm{1}−\sqrt{\mathrm{3}}\:\:<\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}=\mathrm{2}+\sqrt{\mathrm{3}}\:−\mathrm{1}=\mathrm{1}+\sqrt{\mathrm{3}}\:>\mathrm{1}\:\left({to}\:{eliminate}\:{from}\:{residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:{but}\:\varphi\left({z}\right)=\:\frac{−\mathrm{2}{i}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$$\Rightarrow\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\:\frac{−\mathrm{2}{i}}{{z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} }=\:\frac{−\mathrm{2}{i}}{\mathrm{2}\sqrt{\mathrm{3}}}\:\:\:=\frac{−{i}}{\sqrt{\mathrm{3}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:.\frac{−{i}}{\sqrt{\mathrm{3}}}=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}\:}\:\Rightarrow\:\:{I}=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:. \\ $$

Answered by Joel578 last updated on 12/Mar/18

I = ∫_0 ^(2π) (dx/(2 + cos x)) = 2 ∫_0 ^π (dx/(2 + cos x)) t = tan ((x/2)) → dt = ((1 + t^2 )/2) dx cos x = ((1 − t^2 )/(1 + t^2 )) I = 2 ∫_0 ^∞ ((2/(1 + t^2 ))/(((2 + 2t^2 )/(1 + t^2 )) + ((1 − t^2 )/(1 + t^2 )))) dt = 4 lim_(n→∞) (∫_0 ^n (dt/(3 + t^2 ))) = 4 lim_(n→∞) [(1/(√3))tan^(−1) ((t/(√3)))]_0 ^n = 4 lim_(n→∞) ( (1/(√3))tan^(−1) ((n/(√3))) − (1/(√3))tan^(−1) (0) ) = 4 . (1/(√3)) tan^(−1) (∞) I = 4 . (1/(√3)) . (π/2) = ((2π)/(√3))

$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dx}}{\mathrm{2}\:+\:\mathrm{cos}\:{x}}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\frac{{dx}}{\mathrm{2}\:+\:\mathrm{cos}\:{x}} \\ $$$$ \\ $$$${t}\:=\:\mathrm{tan}\:\left(\frac{{x}}{\mathrm{2}}\right)\:\:\rightarrow\:\:{dt}\:=\:\frac{\mathrm{1}\:+\:{t}^{\mathrm{2}} }{\mathrm{2}}\:{dx} \\ $$$$\mathrm{cos}\:{x}\:=\:\frac{\mathrm{1}\:−\:{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} } \\ $$$$ \\ $$$${I}\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\frac{\frac{\mathrm{2}}{\mathrm{1}\:+\:{t}^{\mathrm{2}} }}{\frac{\mathrm{2}\:+\:\mathrm{2}{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }\:+\:\frac{\mathrm{1}\:−\:{t}^{\mathrm{2}} }{\mathrm{1}\:+\:{t}^{\mathrm{2}} }}\:{dt} \\ $$$$\:\:\:\:=\:\mathrm{4}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\int_{\mathrm{0}} ^{{n}} \:\frac{{dt}}{\mathrm{3}\:+\:{t}^{\mathrm{2}} }\right) \\ $$$$\:\:\:\:=\:\mathrm{4}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left[\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{t}}{\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{{n}} \\ $$$$\:\:\:\:=\:\mathrm{4}\:\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\left(\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \left(\frac{{n}}{\sqrt{\mathrm{3}}}\right)\:−\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\mathrm{tan}^{−\mathrm{1}} \:\left(\mathrm{0}\right)\:\right) \\ $$$$\:\:\:\:=\:\mathrm{4}\:.\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \left(\infty\right) \\ $$$${I}\:\:=\:\mathrm{4}\:.\:\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:.\:\frac{\pi}{\mathrm{2}}\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}} \\ $$

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