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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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