calculate-0-2pi-dx-1-2cosx-

Question Number 32483 by prof Abdo imad last updated on 25/Mar/18

c a l c u l a t e \int_{0}^{2 π} \frac{d x}{1 + 2 c o s x} .

Commented by prof Abdo imad last updated on 26/Mar/18

l e t p u t I = \int_{0}^{2 π} \frac{d x}{1 + 2 c o s x} t h e c h . x = π + t g i v e

I = \int_{- π}^{π} \frac{d t}{1 - 2 c o s t} = 2 \int_{0}^{π} \frac{d t}{1 - 2 c o s t} a n d t h e

c h . t a n (\frac{t}{2}) = u h i v e

I = 2 \int_{0}^{+ \infty} \frac{1}{1 - 2 \frac{1 - u^{2}}{1 + u^{2}}} \frac{2 d u}{1 + u^{2}}

= 4 \int_{0}^{\infty} \frac{d u}{1 + u^{2} - 2 + 2 u^{2}} = 4 \int_{0}^{\infty} \frac{d u}{3 u^{2} - 1}

= 4 \int_{0}^{\infty} \frac{d u}{(\sqrt{3} u + 1) (\sqrt{3} - 1)}

= 2 \int_{0}^{\infty} (\frac{1}{\sqrt{3} t - 1} - \frac{1}{\sqrt{3} t + 1}) d u

= \frac{2}{\sqrt{3}} {[l n ∣ \frac{\sqrt{3} t - 1}{\sqrt{3} t + 1} ∣]}_{0}^{+ \infty} = 0

I = 0

Commented by prof Abdo imad last updated on 26/Mar/18

m e t h o d o f r e s i d u s c h . e^{i x} = z g i v e

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

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

= \int_{∣ z ∣= 1} \frac{- i d z}{z^{2} + z + 1} l e t i m 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{- i d z}{z^{2} + z + 1} . p o l e s o f φ ?

z^{2} + z + 1 = 0 \Rightarrow Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2}

z_{1} = \frac{- 1 + i \sqrt{3}}{2} = j a n d z_{2} = \frac{- 1 - i \sqrt{3}}{2} = \bar{j}

∣ z_{1} ∣= 1 a n d ∣ z_{2} ∣= 1 s o

\int_{∣ z ∣= 1} φ (z) d z = 2 i π (R e s (φ, z_{1}) + R e s (φ, z_{2}))

b u t w e h a v e φ (z) = \frac{- i}{(z - z_{1}) (z - z_{2})}

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

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

\int_{∣ z ∣= 1} φ (z) d z = 2 i π (\frac{- 1}{\sqrt{3}} + \frac{1}{\sqrt{3}}) = 0 \Rightarrow I = 0