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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
$let I = ∫_0 ^(2π) ((cosβ)/(10 +8sinβ)) dβ changement e^(iβ) =z give I = ∫_(∣z∣=1) (((z+z^(−1) )/2)/(10 +8 ((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((z+z^(−1) )/(iz(20 +((z−z^(−1) )/i))))dz = ∫_(∣z∣=1) ((z+z^(−1) )/(20iz +z^2 −1)) dz = ∫_(∣z∣=1) ((z+z^(−1) )/({ z^2 +20iz −1}))dz let ϕ(z) = ((z+z^(−1) )/({z^2 +20iz −1})) poles of ϕ? roots of z^2 +20iz −1 Δ^′ =(10i)^2 +1 =−99 =(i(√(99)))^2 ⇒z_1 =−10i +i(√(99))=i(−10 +(√)99) z_2 =−10i −i(√(99))=i(−10−(√(99))) ∣z_1 ∣−1 =(√((−10 +(√(99)))^2 )) −1 = 10−(√(99))−1=9−(√(99))<0 ∣z_2 ∣ −1 =(√((−10−(√(99)))^2 )) −1=10 +(√(99))−1 =9+(√(99))>0( out of circle)⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ {Res(ϕ,z_1 )} Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) = ((z_1 +z_1 ^(−1) )/((z_1 −z_2 ))) = ((i(−10+(√(99)))−i(−10+(√(99))))/(2i(√(99)))) =0 ⇒ I =0$
$l e t I = \int_{0}^{2 π} \frac{c o s β}{10 + 8 s i n β} d β c h a n g e m e n t e^{i β} = z g i v e$ $I = \int_{∣ z ∣= 1} \frac{\frac{z + z^{- 1}}{2}}{10 + 8 \frac{z - z^{- 1}}{2 i}} \frac{d z}{i z} = \int_{∣ z ∣= 1} \frac{z + z^{- 1}}{i z (20 + \frac{z - z^{- 1}}{i})} d z$ $= \int_{∣ z ∣= 1} \frac{z + z^{- 1}}{20 i z + z^{2} - 1} d z = \int_{∣ z ∣= 1} \frac{z + z^{- 1}}{{z^{2} + 20 i z - 1}} d z l e t$ $φ (z) = \frac{z + z^{- 1}}{{z^{2} + 20 i z - 1}} p o l e s o f φ ? r o o t s o f z^{2} + 20 i z - 1$ $Δ^{^{'}} = {(10 i)}^{2} + 1 = - 99 = {(i \sqrt{99})}^{2} \Rightarrow z_{1} = - 10 i + i \sqrt{99} = i (- 10 + \sqrt{} 99)$ $z_{2} = - 10 i - i \sqrt{99} = i (- 10 - \sqrt{99})$ $∣ z_{1} ∣ - 1 = \sqrt{{(- 10 + \sqrt{99})}^{2}} - 1 = 10 - \sqrt{99} - 1 = 9 - \sqrt{99} < 0$ $∣ z_{2} ∣ - 1 = \sqrt{{(- 10 - \sqrt{99})}^{2}} - 1 = 10 + \sqrt{99} - 1 = 9 + \sqrt{99} > 0 (o u t o f c i r c l e) \Rightarrow$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π {R e s (φ, z_{1})}$ $R e s (φ, z_{1}) = {l i m}_{z \to z_{1}} (z - z_{1}) φ (z) = \frac{z_{1} + {\overset{- 1}{z}}_{1}}{(z_{1} - z_{2})} = \frac{i (- 10 + \sqrt{99}) - i (- 10 + \sqrt{99})}{2 i \sqrt{99}}$ $= 0 \Rightarrow I = 0$
Commented by Tawa1 last updated on 12/Sep/18

$God bless you sir .$
Commented by math khazana by abdo last updated on 12/Sep/18

$y o u a r e w e l c o m e s i r .$
	Answered by MrW3 last updated on 12/Sep/18

	$\int_{0}^{2 π} \frac{\cos β}{10 + 8 \sin β} d β$ $= \int_{0}^{2 π} \frac{1}{10 + 8 \sin β} d \sin β$ $= \frac{1}{8} \int_{0}^{2 π} \frac{1}{10 + 8 \sin β} d (10 + 8 \sin β)$ $= \frac{1}{8} {[\ln (10 + 8 \sin β)]}_{0}^{2 π}$ $= \frac{1}{8} [\ln 10 - \ln 10]$ $= 0$
	Commented by Tawa1 last updated on 12/Sep/18

	$God bless you sir$
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