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Question Number 147992 by Sozan last updated on 24/Jul/21

Commented by Sozan last updated on 25/Jul/21

$w h o i s c a n s o l v e t h i s p l e a s e ?$
	Answered by mathmax by abdo last updated on 25/Jul/21
	$Ψ=∫_0 ^(2π) ((sin(2x))/(3+4sinx)) ⇒Ψ=∫_0 ^(2π) ((2sinx.cosx)/(3+4sinx))dx =_(e^(ix) =z) ∫_(∣z∣=1) ((2((z−z^(−1) )/(2i)).((z+z^(−1) )/2))/(3+4((z−z^(−1) )/(2i)))) (dz/(iz)) =∫_(∣z∣=1) ((z^2 −z^(−2) )/(iz(6i+4z−4z^(−1) )))dz =−i∫_(∣z∣=1) ((z^2 −z^(−2) )/(6iz+4z^2 −4))dz =−(i/2)∫_(∣z∣=1) ((z^2 −z^(−2) )/(2z^2 +3iz−2))dz let ϕ(z)=((z^2 −z^(−2) )/(2z^2 +3iz−2)) poles? Δ=−9−4(−4) =16−9=7 ⇒z_1 =((−3i+(√7))/4) z_2 =((−3i−(√7))/4) ∣z_1 ∣−1 =(1/4)(√(9+7))−1 =0 ⇒∣z_1 ∣=1 z_1 z_2 =1 ⇒∣z_2 ∣=1 residus ⇒∫_(∣z∣=1) ϕ(z)dz=2iπ{Res(ϕ,z_1 )+Res(ϕ,z_2 )} but Re(ϕ,z_2 )=−Res(ϕ,z_1 )⇒∫_(∣z∣=1) ϕ(z)dz=0 ⇒Ψ=0 another way Ψ=∫_0 ^π ()dx +∫_π ^(2π) ()dx =_(x=π+α) ∫_0 ^π ((sin2x)/(3+4sinx))dx +∫_0 ^π ((sin(2α))/(3−4sinα))dα =∫_0 ^π sin(2x){(1/(3+4sinx))+(1/(3−4sinx))}dx =∫_0 ^π ((6sin(2x))/(9−16sin^2 x))dx =∫_0 ^π ((6sin(2x))/(9−16(1−cos^2 x)))dx =∫_0 ^π ((6sin(2x))/(16cos^2 x−7))dx =∫_0 ^π ((6sin(2x))/(16.((1+cos(2x))/2)−7))dx =∫_0 ^π ((6sin(2x))/(1+8cos(2x)))dx =(6/(16))∫_0 ^π ((16sin(2x))/(8cos(2x)+1))dx =(3/8)[ln∣8cos(2x)+1∣]_0 ^π =(3/8){ln9−log9}=0$
	$Ψ = \int_{0}^{2 π} \frac{\sin (2 x)}{3 + 4 sinx} \Rightarrow Ψ = \int_{0}^{2 π} \frac{2 sinx . cosx}{3 + 4 sinx} dx$ $=_{e^{ix} = z} \int_{∣ z ∣= 1} \frac{2 \frac{z - z^{- 1}}{2 i} . \frac{z + z^{- 1}}{2}}{3 + 4 \frac{z - z^{- 1}}{2 i}} \frac{dz}{iz}$ $= \int_{∣ z ∣= 1} \frac{z^{2} - z^{- 2}}{iz (6 i + 4 z - {4 z}^{- 1})} dz = - i \int_{∣ z ∣= 1} \frac{z^{2} - z^{- 2}}{6 iz + {4 z}^{2} - 4} dz$ $= - \frac{i}{2} \int_{∣ z ∣= 1} \frac{z^{2} - z^{- 2}}{{2 z}^{2} + 3 iz - 2} dz let φ (z) = \frac{z^{2} - z^{- 2}}{{2 z}^{2} + 3 iz - 2} poles ?$ $Δ = - 9 - 4 (- 4) = 16 - 9 = 7 \Rightarrow z_{1} = \frac{- 3 i + \sqrt{7}}{4}$ $z_{2} = \frac{- 3 i - \sqrt{7}}{4}$ $∣ z_{1} ∣ - 1 = \frac{1}{4} \sqrt{9 + 7} - 1 = 0 \Rightarrow∣ z_{1} ∣= 1$ $z_{1} z_{2} = 1 \Rightarrow∣ z_{2} ∣= 1 residus \Rightarrow \int_{∣ z ∣= 1} φ (z) dz = 2 i π {Res (φ, z_{1}) + Res (φ, z_{2})}$ $but Re (φ, z_{2}) = - Res (φ, z_{1}) \Rightarrow \int_{∣ z ∣= 1} φ (z) dz = 0 \Rightarrow Ψ = 0$ $another way Ψ = \int_{0}^{π} () dx + \int_{π}^{2 π} () dx =_{x = π + α} \int_{0}^{π} \frac{\sin 2 x}{3 + 4 sinx} dx + \int_{0}^{π} \frac{\sin (2 α)}{3 - 4 \sin α} d α$ $= \int_{0}^{π} \sin (2 x) {\frac{1}{3 + 4 sinx} + \frac{1}{3 - 4 sinx}} dx$ $= \int_{0}^{π} \frac{6 \sin (2 x)}{9 - {16 \sin}^{2} x} dx = \int_{0}^{π} \frac{6 \sin (2 x)}{9 - 16 (1 - \cos^{2} x)} dx$ $= \int_{0}^{π} \frac{6 \sin (2 x)}{{16 \cos}^{2} x - 7} dx = \int_{0}^{π} \frac{6 \sin (2 x)}{16 . \frac{1 + \cos (2 x)}{2} - 7} dx$ $= \int_{0}^{π} \frac{6 \sin (2 x)}{1 + 8 \cos (2 x)} dx = \frac{6}{16} \int_{0}^{π} \frac{16 \sin (2 x)}{8 \cos (2 x) + 1} dx$ $= \frac{3}{8} {[\ln ∣ 8 \cos (2 x) + 1 ∣]}_{0}^{π} = \frac{3}{8} {\ln 9 - \log 9} = 0$
	Commented by tabata last updated on 25/Jul/21

	$m s r m a t h m a x b y a b d o c a n y o u h e l p m e i n q u e s t i o n 148064 p l e s e$
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