find-the-value-of-I-0-pi-dx-2-cosx-

Question Number 35729 by abdo mathsup 649 cc last updated on 22/May/18

f i n d t h e v a l u e o f I = \int_{0}^{π} \frac{d x}{2 - c o s x}

Commented by abdo mathsup 649 cc last updated on 23/May/18

c h a n g e m e n t x = 2 t g i v e

I = \int_{0}^{2 π} \frac{2 d t}{2 - c o s (2 t)} t h e n c h a n g e m e n t e^{i t} = z g i v e

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

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

\int_{∣ z ∣ = 1} \frac{- 4 i z d z}{4 z - z^{3} - z^{- 1}} = \int_{∣ z ∣= 1} \frac{- 4 i z d z}{4 z^{2} - z^{4} - 1}

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

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

o f z^{4} - 4 z^{2} + 1 = 0 \Rightarrow t^{2} - 4 t + 1 = 0 (t = z^{2})

Δ^{^{'}} = 4 - 1 = 3 \Rightarrow t_{1} = 2 + \sqrt{3} a n d t_{2} = 2 - \sqrt{3}

\Rightarrow φ (z) = \frac{4 i z}{(z^{2} - (2 + \sqrt{3})) (z^{2} - (2 - \sqrt{3}))}

= \frac{4 i z}{(z - \sqrt{2 + \sqrt{3}}) (z + \sqrt{2 + \sqrt{3}) (z - \sqrt{2 - \sqrt{3}}}) (z + \sqrt{2 - \sqrt{3}}})

t h e p o l e s o f φ a r e \bar{+} \sqrt{2 + \sqrt{3}} a n d \bar{+} \sqrt{2 - \sqrt{3}}

\int_{∣ z ∣= 1} φ (z) d z = 2 i π R e s (φ, \sqrt{2 - \sqrt{3}})

R e s (φ, \sqrt{2 - \sqrt{3}}) = \frac{4 i \sqrt{2 - \sqrt{3}}}{2 \sqrt{2 - \sqrt{3}} (2 - \sqrt{3} - 2 - \sqrt{3})}

= \frac{4 i}{- 4 \sqrt{3}} = \frac{- i}{\sqrt{3}}

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

\Rightarrow I = \frac{2 π}{\sqrt{3}} .

Answered by MJS last updated on 23/May/18

\underset{0}{\int^{π}} \frac{d x}{2 - \cos (x)} =

t = \tan (\frac{x}{2}) \to d x = \frac{2 d t}{1 + t^{2}}; \cos (x) = \frac{1 - t^{2}}{1 + t^{2}}

= \underset{0}{\int^{\infty}} \frac{2 d t}{3 t^{2} + 1} =

u = \sqrt{3} t \to d t = \frac{\sqrt{3}}{3} d u

= \frac{2 \sqrt{3}}{3} \underset{0}{\int^{\infty}} \frac{d u}{u^{2} + 1} = {[\frac{2 \sqrt{3}}{3} \arctan (u)]}_{0}^{\infty} =

= {[\frac{2 \sqrt{3}}{3} \arctan (\sqrt{3} t)]}_{0}^{\infty} =

= {[\frac{2 \sqrt{3}}{3} \arctan (\sqrt{3} \tan (\frac{x}{2}))]}_{0}^{π} = \frac{2 π \sqrt{3}}{3}

Facebook Tweet Pin