find-0-2pi-cos-n-2-3cos-d-n-from-N-

Question Number 29856 by abdo imad last updated on 13/Feb/18

f i n d \int_{0}^{2 π} \frac{c o s (n θ)}{2 + 3 c o s θ} d θ . n f r o m N .

Commented by prof Abdo imad last updated on 18/Feb/18

l e t p u t I = \int_{0}^{2 π} \frac{c o s (n θ)}{2 + 3 c o s θ} d θ

I = R e (\int_{0}^{2 π} \frac{e^{i n θ}}{2 + c o s θ} d θ) = R e (w_{n}) c h . e^{i θ} = z g i v e

w_{n} = \int_{∣ z ∣= 1} \frac{z^{n}}{2 + \frac{z + z^{- 1}}{2}} \frac{d z}{i z}

= \int_{∣ z ∣= 1} \frac{2 z^{n}}{i z (4 + z + z^{- 1})} d z

= \int_{∣ z ∣= 1} \frac{- 2 {i z}^{n}}{4 z + z^{2} + 1} d z = \int_{∣ z ∣= 1} \frac{- 2 i z^{n}}{z^{2} + 4 z + 1} d z l e t

i n t r o d u c e t h e c o m p l e x f u n c t i o n

φ (z) = \frac{- 2 {i z}^{n}}{z^{2} + 4 z + 1} p o l e s o f φ ?

Δ^{^{'}} = 2^{2} - 1 = 3 \Rightarrow z_{1} = - 2 + \sqrt{3} a n d z_{2} = - 2 - \sqrt{3}

∣ z_{1} ∣ - 1 = 2 - \sqrt{3} - 1 = 1 - \sqrt{3} < 1 a n d

∣ z_{2} ∣ - 1 = 2 + \sqrt{3} - 1 = 1 + \sqrt{3} > 1 (t o e l i m i n a t e f r o m

r e s i d u s) s o

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, z_{1}) b u t w e h a v e

φ (z) = \frac{- 2 i z^{n}}{(z - z_{1}) (z - z_{2})} \Rightarrow

R e s (φ, z_{1}) = \frac{- 2 i z_{1}^{n}}{z_{1} - z_{2}} = \frac{- 2 i {(- 2 + \sqrt{3})}^{n}}{2 \sqrt{3}}

= \frac{- i {(- 2 + \sqrt{3})}^{n}}{\sqrt{3}}

\int_{∣ z ∣= 1} φ (z) d z = \frac{2 π}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n}

R e (\int \dots . .) = \frac{2 π}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n} \Rightarrow

I = \frac{2 π}{\sqrt{3}} {(- 2 + \sqrt{3})}^{n} .