solve-the-integral-with-Residue-theorem-0-2pi-3-d-9-sin-2-

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 = \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