calculate ∫_0 ^(2π) (dx/(1+2cosx)) .


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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

$m e t h o d o f r e s i d u s c h . e^{i x} = z g i v e$ $I = \int_{∣ z ∣= 1} \frac{1}{1 + 2 \frac{z + z^{- 1}}{2}} \frac{d z}{i z}$ $= \int_{∣ z ∣= 1} \frac{d z}{i z (1 + z + z^{- 1})} = \int_{∣ z ∣= 1} \frac{- i d z}{z + z^{2} + 1}$ $= \int_{∣ z ∣= 1} \frac{- i d z}{z^{2} + z + 1} l e t i m t r o d u c e t h e c o m p l e x$ $f u n c t i o n φ (z) = \frac{- i d z}{z^{2} + z + 1} . p o l e s o f φ ?$ $z^{2} + z + 1 = 0 \Rightarrow Δ = 1 - 4 = - 3 = {(i \sqrt{3})}^{2}$ $z_{1} = \frac{- 1 + i \sqrt{3}}{2} = j a n d z_{2} = \frac{- 1 - i \sqrt{3}}{2} = \bar{j}$ $∣ z_{1} ∣= 1 a n d ∣ z_{2} ∣= 1 s o$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π (R e s (φ, z_{1}) + R e s (φ, z_{2}))$ $b u t w e h a v e φ (z) = \frac{- i}{(z - z_{1}) (z - z_{2})}$ $R e s (φ, z_{1}) = \frac{- i}{z_{1} - z_{2}} = \frac{- i}{i \sqrt{3}} = \frac{- 1}{\sqrt{3}}$ $R e s (φ, z_{2}) = \frac{- i}{z_{2} - z_{1}} = \frac{- i}{- i \sqrt{3}} = \frac{1}{\sqrt{3}}$ $\int_{∣ z ∣= 1} φ (z) d z = 2 i π (\frac{- 1}{\sqrt{3}} + \frac{1}{\sqrt{3}}) = 0 \Rightarrow I = 0$
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