Menu Close

find-the-value-of-I-0-pi-dx-2-cosx-

Tinku Tara 0 Comments
Pin




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

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *