solve the integral with Residue theorem. ∫_0 ^(2π) ((3 dθ)/(9 +sin^2 θ))


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Question Number 75435 by Crabby89p13 last updated on 11/Dec/19

$s o l v e t h e i n t e g r a l w i t h R e s i d u e t h e o r e m .$ $\underset{0}{\int^{2 π}} \frac{3 d θ}{9 + \sin^{2} θ}$
Commented by mathmax by abdo last updated on 11/Dec/19
$I =∫_0 ^(2π) ((3dθ)/(9+sin^2 θ)) ⇒I =∫_0 ^(2π) ((3dθ)/(9+((1−cos(2θ))/2))) =∫_0 ^(2π) ((6dθ)/(18+1−cos(2θ))) =_(2θ =t) ∫_0 ^(4π) (6/(19−cost))(dt/2) =3 ∫_0 ^(4π) (dt/(19−cost)) =3{ ∫_0 ^(2π) (dt/(19−cost)) +∫_(2π) ^(4π) (dt/(19−cost))} but ∫_(2π) ^(4π) (dt/(19−cost)) =_(t=2π +u) ∫_0 ^(2π) (du/(19−cosu)) ⇒I =6 ∫_0 ^(2π) (dt/(19−cost)) we have ∫_0 ^(2π) (dt/(19−cost)) =_(e^(it) =z) ∫_(∣z∣=1) (1/(19−((z+z^(−1) )/2)))(dz/(iz)) =∫_(∣z∣=1) ((2dz)/(iz{38−z−z^(−1) })) =∫_(∣z∣=1) ((−2idz)/(38z−z^2 −1)) =∫_(∣z∣=1) ((2idz)/(z^2 −38z +1)) let W(z)=((2i)/(z^2 −38z +1)) poles of W? z^2 −38z +1=0→Δ^′ =19^2 −1 =360 z_1 =19+6(√(10)) and z_2 =19−6(√(10)) ∣z_1 ∣−1 =18+6(√(10))>0 (z_1 is out of circle) ∣z_2 ∣−1 =18−6(√(10)) <0 ⇒ ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_2 ) =2iπ×((2i)/((z_2 −z_1 ))) =((−4π)/(−12(√(10)))) =(π/(3(√(10)))) ⇒ I =6×(π/(3(√(10)))) =((2π)/(√(10)))$
$I = \int_{0}^{2 π} \frac{3 d θ}{9 + {s i n}^{2} θ} \Rightarrow I = \int_{0}^{2 π} \frac{3 d θ}{9 + \frac{1 - c o s (2 θ)}{2}} = \int_{0}^{2 π} \frac{6 d θ}{18 + 1 - c o s (2 θ)}$ $=_{2 θ = t} \int_{0}^{4 π} \frac{6}{19 - c o s t} \frac{d t}{2} = 3 \int_{0}^{4 π} \frac{d t}{19 - c o s t}$ $= 3 {\int_{0}^{2 π} \frac{d t}{19 - c o s t} + \int_{2 π}^{4 π} \frac{d t}{19 - c o s t}} b u t$ $\int_{2 π}^{4 π} \frac{d t}{19 - c o s t} =_{t = 2 π + u} \int_{0}^{2 π} \frac{d u}{19 - c o s u} \Rightarrow I = 6 \int_{0}^{2 π} \frac{d t}{19 - c o s t}$ $w e h a v e \int_{0}^{2 π} \frac{d t}{19 - c o s t} =_{e^{i t} = z} \int_{∣ z ∣= 1} \frac{1}{19 - \frac{z + z^{- 1}}{2}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{2 d z}{i z {38 - z - z^{- 1}}} = \int_{∣ z ∣= 1} \frac{- 2 i d z}{38 z - z^{2} - 1}$ $= \int_{∣ z ∣= 1} \frac{2 i d z}{z^{2} - 38 z + 1} l e t W (z) = \frac{2 i}{z^{2} - 38 z + 1} p o l e s o f W ?$ $z^{2} - 38 z + 1 = 0 \to Δ^{^{'}} = 19^{2} - 1 = 360$ $z_{1} = 19 + 6 \sqrt{10} a n d z_{2} = 19 - 6 \sqrt{10}$ $∣ z_{1} ∣ - 1 = 18 + 6 \sqrt{10} > 0 (z_{1} i s o u t o f c i r c l e)$ $∣ z_{2} ∣ - 1 = 18 - 6 \sqrt{10} < 0 \Rightarrow$ $\int_{∣ z ∣= 1} W (z) d z = 2 i π R e s (W, z_{2}) = 2 i π \times \frac{2 i}{(z_{2} - z_{1})}$ $= \frac{- 4 π}{- 12 \sqrt{10}} = \frac{π}{3 \sqrt{10}} \Rightarrow I = 6 \times \frac{π}{3 \sqrt{10}} = \frac{2 π}{\sqrt{10}}$
Commented by Crabby89p13 last updated on 11/Dec/19

$w h y n o t u s e d r e s i d u e f o r m u l a f o r s i n θ ?$ $\frac{z - z^{- 1}}{2 j}$
Commented by mathmax by abdo last updated on 11/Dec/19

$w e u s e l i e a r i s a t i o n t o d e c r e a s e t h e d e g r e .$
Commented by Crabby89p13 last updated on 12/Dec/19

$o h i c . . t h a n k s$
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