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Question Number 35729 by abdo mathsup 649 cc last updated on 22/May/18

find the value of   I =∫_0 ^π     (dx/(2−cosx))

$${find}\:{the}\:{value}\:{of}\:\:\:{I}\:=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}−{cosx}} \\ $$

Commented by abdo mathsup 649 cc last updated on 23/May/18

changement x=2t give   I = ∫_0 ^(2π)     ((2dt)/(2 −cos(2t)))  then changement e^(it)  =z give  I  = ∫_(∣z∣=1)     (2/(2− ((z^2  +z^(−2) )/2))) (dz/(iz))   =∫_(∣z∣=1)    ((4dz)/(iz( 4 −z^2  −z^(−2) )))   ∫_(∣z∣ =1)        ((−4iz dz)/(4z −z^3  −z^(−1) )) = ∫_(∣z∣=1)    ((−4izdz)/(4z^2  −z^4  −1))   = ∫_(∣z∣ =1)     ((4i zdz)/(z^4  −4z^2  +1)) .let consider  ϕ(z) = ((4iz)/(z^4  −4z^2  +1)) .poles of ϕ? let find the roots  of  z^4  −4z^2  +1 =0 ⇒ t^2  −4t +1 =0  ( t=z^2 )  Δ^′  = 4−1 =3 ⇒ t_1 = 2+(√3)   and t_2 = 2−(√( 3))  ⇒  ϕ(z) = ((4iz)/((z^2 −(2+(√3)))(z^2  −(2−(√3)))))  =  ((4iz)/((z −(√(2+(√3))) )(z +(√(2+(√3))( z−(√(2−(√3))))))(z+(√(2−(√3))))))  the poles of ϕ are  +^− (√(2+(√3)))   and +^− (√(2−(√3)))  ∫_(∣z∣=1)    ϕ(z)dz =2iπ Res(ϕ, (√(2−(√3))) )  Res(ϕ,(√(2−(√3))) ) =  ((4i(√(2−(√3))))/(2(√(2−(√3)))(2−(√3) −2−(√3))))  = ((4i)/(−4(√3))) =((−i)/(√3))  ∫_(∣z∣=1)    ϕ(z)dz =  2iπ  ((−i)/(√3)) =  ((2π)/(√3))  ⇒  I  = ((2π)/(√3))  .

$${changement}\:{x}=\mathrm{2}{t}\:{give}\: \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}\:−{cos}\left(\mathrm{2}{t}\right)}\:\:{then}\:{changement}\:{e}^{{it}} \:={z}\:{give} \\ $$$${I}\:\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}−\:\frac{{z}^{\mathrm{2}} \:+{z}^{−\mathrm{2}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}}\: \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{4}{dz}}{{iz}\left(\:\mathrm{4}\:−{z}^{\mathrm{2}} \:−{z}^{−\mathrm{2}} \right)}\: \\ $$$$\int_{\mid{z}\mid\:=\mathrm{1}} \:\:\:\:\:\:\:\frac{−\mathrm{4}{iz}\:{dz}}{\mathrm{4}{z}\:−{z}^{\mathrm{3}} \:−{z}^{−\mathrm{1}} }\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{4}{izdz}}{\mathrm{4}{z}^{\mathrm{2}} \:−{z}^{\mathrm{4}} \:−\mathrm{1}}\: \\ $$$$=\:\int_{\mid{z}\mid\:=\mathrm{1}} \:\:\:\:\frac{\mathrm{4}{i}\:{zdz}}{{z}^{\mathrm{4}} \:−\mathrm{4}{z}^{\mathrm{2}} \:+\mathrm{1}}\:.{let}\:{consider} \\ $$$$\varphi\left({z}\right)\:=\:\frac{\mathrm{4}{iz}}{{z}^{\mathrm{4}} \:−\mathrm{4}{z}^{\mathrm{2}} \:+\mathrm{1}}\:.{poles}\:{of}\:\varphi?\:{let}\:{find}\:{the}\:{roots} \\ $$$${of}\:\:{z}^{\mathrm{4}} \:−\mathrm{4}{z}^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{0}\:\Rightarrow\:{t}^{\mathrm{2}} \:−\mathrm{4}{t}\:+\mathrm{1}\:=\mathrm{0}\:\:\left(\:{t}={z}^{\mathrm{2}} \right) \\ $$$$\Delta^{'} \:=\:\mathrm{4}−\mathrm{1}\:=\mathrm{3}\:\Rightarrow\:{t}_{\mathrm{1}} =\:\mathrm{2}+\sqrt{\mathrm{3}}\:\:\:{and}\:{t}_{\mathrm{2}} =\:\mathrm{2}−\sqrt{\:\mathrm{3}} \\ $$$$\Rightarrow\:\:\varphi\left({z}\right)\:=\:\frac{\mathrm{4}{iz}}{\left({z}^{\mathrm{2}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right)\left({z}^{\mathrm{2}} \:−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\right)} \\ $$$$\left.=\:\:\frac{\mathrm{4}{iz}}{\left({z}\:−\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\right)\left({z}\:+\sqrt{\left.\mathrm{2}+\sqrt{\mathrm{3}}\right)\left(\:{z}−\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right.}\right)\left({z}+\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\right.}\right) \\ $$$${the}\:{poles}\:{of}\:\varphi\:{are}\:\:\overset{−} {+}\sqrt{\mathrm{2}+\sqrt{\mathrm{3}}}\:\:\:{and}\:\overset{−} {+}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\:\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right) \\ $$$${Res}\left(\varphi,\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\:\right)\:=\:\:\frac{\mathrm{4}{i}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}}{\mathrm{2}\sqrt{\mathrm{2}−\sqrt{\mathrm{3}}}\left(\mathrm{2}−\sqrt{\mathrm{3}}\:−\mathrm{2}−\sqrt{\mathrm{3}}\right)} \\ $$$$=\:\frac{\mathrm{4}{i}}{−\mathrm{4}\sqrt{\mathrm{3}}}\:=\frac{−{i}}{\sqrt{\mathrm{3}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\varphi\left({z}\right){dz}\:=\:\:\mathrm{2}{i}\pi\:\:\frac{−{i}}{\sqrt{\mathrm{3}}}\:=\:\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}} \\ $$$$\Rightarrow\:\:{I}\:\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{3}}}\:\:. \\ $$

Answered by MJS last updated on 23/May/18

∫_0 ^π (dx/(2−cos(x) ))=            t=tan((x/2)) → dx=((2dt)/(1+t^2 )); cos(x)=((1−t^2 )/(1+t^2 ))  =∫_0 ^∞ ((2dt)/(3t^2 +1))=            u=(√3)t → dt=((√3)/3)du  =((2(√3))/3)∫_0 ^∞ (du/(u^2 +1))=[((2(√3))/3)arctan(u)]_0 ^∞ =  =[((2(√3))/3)arctan((√3)t)]_0 ^∞ =  =[((2(√3))/3)arctan((√3)tan((x/2)))]_0 ^π =((2π(√3))/3)

$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{dx}}{\mathrm{2}−\mathrm{cos}\left({x}\right)\:}= \\ $$$$\:\:\:\:\:\:\:\:\:\:{t}=\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:\rightarrow\:{dx}=\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} };\:\mathrm{cos}\left({x}\right)=\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{\mathrm{2}{dt}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:{u}=\sqrt{\mathrm{3}}{t}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}{du} \\ $$$$=\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\left[\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left({u}\right)\right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\left[\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left(\sqrt{\mathrm{3}}{t}\right)\right]_{\mathrm{0}} ^{\infty} = \\ $$$$=\left[\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\mathrm{arctan}\left(\sqrt{\mathrm{3}}\mathrm{tan}\left(\frac{{x}}{\mathrm{2}}\right)\right)\right]_{\mathrm{0}} ^{\pi} =\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{3}} \\ $$

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