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Question Number 43589 by Tawa1 last updated on 12/Sep/18

Commented by maxmathsup by imad last updated on 12/Sep/18

let I = ∫_0 ^(2π) ((cosβ)/(10 +8sinβ)) dβ changement e^(iβ) =z give I = ∫_(∣z∣=1) (((z+z^(−1) )/2)/(10 +8 ((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((z+z^(−1) )/(iz(20 +((z−z^(−1) )/i))))dz = ∫_(∣z∣=1) ((z+z^(−1) )/(20iz +z^2 −1)) dz = ∫_(∣z∣=1) ((z+z^(−1) )/({ z^2 +20iz −1}))dz let ϕ(z) = ((z+z^(−1) )/({z^2 +20iz −1})) poles of ϕ? roots of z^2 +20iz −1 Δ^′ =(10i)^2 +1 =−99 =(i(√(99)))^2 ⇒z_1 =−10i +i(√(99))=i(−10 +(√)99) z_2 =−10i −i(√(99))=i(−10−(√(99))) ∣z_1 ∣−1 =(√((−10 +(√(99)))^2 )) −1 = 10−(√(99))−1=9−(√(99))<0 ∣z_2 ∣ −1 =(√((−10−(√(99)))^2 )) −1=10 +(√(99))−1 =9+(√(99))>0( out of circle)⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ {Res(ϕ,z_1 )} Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) = ((z_1 +z_1 ^(−1) )/((z_1 −z_2 ))) = ((i(−10+(√(99)))−i(−10+(√(99))))/(2i(√(99)))) =0 ⇒ I =0

$${let}\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}\beta}{\mathrm{10}\:+\mathrm{8}{sin}\beta}\:{d}\beta\:\:\:{changement}\:\:{e}^{{i}\beta} ={z}\:{give} \\ $$$${I}\:\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}{\mathrm{10}\:+\mathrm{8}\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\:\frac{{dz}}{{iz}}\:\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{{z}+{z}^{−\mathrm{1}} }{{iz}\left(\mathrm{20}\:+\frac{{z}−{z}^{−\mathrm{1}} }{{i}}\right)}{dz} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{20}{iz}\:+{z}^{\mathrm{2}} −\mathrm{1}}\:{dz}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{{z}+{z}^{−\mathrm{1}} }{\left\{\:{z}^{\mathrm{2}} \:+\mathrm{20}{iz}\:−\mathrm{1}\right\}}{dz}\:{let}\: \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}+{z}^{−\mathrm{1}} }{\left\{{z}^{\mathrm{2}} \:+\mathrm{20}{iz}\:−\mathrm{1}\right\}}\:\:{poles}\:{of}\:\varphi?\:{roots}\:{of}\:{z}^{\mathrm{2}} \:+\mathrm{20}{iz}\:−\mathrm{1} \\ $$$$\Delta^{'} \:=\left(\mathrm{10}{i}\right)^{\mathrm{2}} +\mathrm{1}\:=−\mathrm{99}\:=\left({i}\sqrt{\mathrm{99}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =−\mathrm{10}{i}\:+{i}\sqrt{\mathrm{99}}={i}\left(−\mathrm{10}\:+\sqrt{}\mathrm{99}\right) \\ $$$${z}_{\mathrm{2}} =−\mathrm{10}{i}\:−{i}\sqrt{\mathrm{99}}={i}\left(−\mathrm{10}−\sqrt{\mathrm{99}}\right) \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\sqrt{\left(−\mathrm{10}\:+\sqrt{\mathrm{99}}\right)^{\mathrm{2}} }\:−\mathrm{1}\:=\:\mathrm{10}−\sqrt{\mathrm{99}}−\mathrm{1}=\mathrm{9}−\sqrt{\mathrm{99}}<\mathrm{0} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}\:=\sqrt{\left(−\mathrm{10}−\sqrt{\mathrm{99}}\right)^{\mathrm{2}} }\:\:−\mathrm{1}=\mathrm{10}\:+\sqrt{\mathrm{99}}−\mathrm{1}\:=\mathrm{9}+\sqrt{\mathrm{99}}>\mathrm{0}\left(\:{out}\:{of}\:{circle}\right)\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\right\} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right)\varphi\left({z}\right)\:=\:\:\frac{{z}_{\mathrm{1}} \:+\overset{−\mathrm{1}} {{z}}_{\mathrm{1}} }{\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\:\frac{{i}\left(−\mathrm{10}+\sqrt{\mathrm{99}}\right)−{i}\left(−\mathrm{10}+\sqrt{\mathrm{99}}\right)}{\mathrm{2}{i}\sqrt{\mathrm{99}}} \\ $$$$=\mathrm{0}\:\:\Rightarrow\:{I}\:=\mathrm{0} \\ $$$$\: \\ $$

Commented by Tawa1 last updated on 12/Sep/18

God bless you sir.

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$

Commented by math khazana by abdo last updated on 12/Sep/18

you are welcome sir.

$${you}\:{are}\:{welcome}\:{sir}. \\ $$

Answered by MrW3 last updated on 12/Sep/18

∫_0 ^( 2π) ((cos β)/(10+8 sin β)) dβ =∫_0 ^( 2π) (1/(10+8 sin β)) d sin β =(1/8)∫_0 ^( 2π) (1/(10+8 sin β)) d (10+8 sin β) =(1/8)[ln (10+8 sin β)]_0 ^(2π) =(1/8)[ln 10−ln 10] =0

$$\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{cos}\:\beta}{\mathrm{10}+\mathrm{8}\:\mathrm{sin}\:\beta}\:{d}\beta \\ $$$$=\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{10}+\mathrm{8}\:\mathrm{sin}\:\beta}\:{d}\:\mathrm{sin}\:\beta \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\int_{\mathrm{0}} ^{\:\mathrm{2}\pi} \frac{\mathrm{1}}{\mathrm{10}+\mathrm{8}\:\mathrm{sin}\:\beta}\:{d}\:\left(\mathrm{10}+\mathrm{8}\:\mathrm{sin}\:\beta\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{ln}\:\left(\mathrm{10}+\mathrm{8}\:\mathrm{sin}\:\beta\right)\right]_{\mathrm{0}} ^{\mathrm{2}\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{8}}\left[\mathrm{ln}\:\mathrm{10}−\mathrm{ln}\:\mathrm{10}\right] \\ $$$$=\mathrm{0} \\ $$

Commented by Tawa1 last updated on 12/Sep/18

God bless you sir

$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$

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