calculate ∫_0 ^(2π) (dθ/((2+cosθ)^2 ))


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Question Number 36200 by prof Abdo imad last updated on 30/May/18

$c a l c u l a t e \int_{0}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}$
Commented by prof Abdo imad last updated on 31/May/18
$letut I = ∫_0 ^(2π) (dθ/((2+cosθ)^2 )) I = ∫_0 ^π (dθ/((2 +cosθ)^2 )) + ∫_π ^(2π) (dθ/((2+cosθ)^2 )) =I_1 + I_2 changement tan((θ/2))=t give I_1 = ∫_0 ^∞ (1/((2 + ((1−t^2 )/(1+t^2 )))^2 )) ((2dt)/(1+t^2 )) = 2 ∫_0 ^∞ (dt/((1+t^2 )(((2 +2t^2 +1−t^2 )/(1+t^2 )))^2 )) = 2 ∫_0 ^∞ (((1+t^2 )^2 )/((1+t^2 )( 3+t^2 )^2 ))dt = 2 ∫_0 ^∞ ((t^2 +1)/(( t^2 +3)^2 ))dt =∫_(−∞) ^(+∞) ((t^2 +1)/((t^2 +3)^2 ))dt let consider the complex function ϕ(z) = ((z^2 +1)/((z^2 +3)^2 )) ϕ(z) = ((z^2 +1)/((z−i(√3))^2 (z +i(√3))^2 )) the poles of ϕ are i(√3) and −i(√3)(doubles) Residus theorem give ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i(√3)) Res(ϕ,i(√3)) =lim_(z→i(√3)) (1/((2−1)!)){ (z−i(√3))^2 ϕ(z)}^((1)) =lim_(z→i(√3)) { ((z^2 +1)/((z +i(√3))^2 ))}^((1)) lim_(z→i(√3) ) { ((2z(z +i(√3))^2 −(z^2 +1)2(z+i(√3)))/((z +i(√3))^4 ))} =lim_(z→i(√3)) { ((2z(z +i(√3)) −2(z^2 +1))/((z +i(√3))^3 ))} =lim_(z→i(√3)) { ((2z^2 + 2iz (√3) −2z^2 −2)/((z+i(√3))^3 ))} = ((2i(i(√3))(√3) −2)/((2i(√3))^3 )) = ((−8)/(−8i3(√3))) = (1/(i3(√3))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ (1/(i3(√3))) = ((2π)/(3(√3))) = I_1$
$l e t u t I = \int_{0}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}$ $I = \int_{0}^{π} \frac{d θ}{{(2 + c o s θ)}^{2}} + \int_{π}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}} = I_{1} + I_{2}$ $c h a n g e m e n t t a n (\frac{θ}{2}) = t g i v e$ $I_{1} = \int_{0}^{\infty} \frac{1}{{(2 + \frac{1 - t^{2}}{1 + t^{2}})}^{2}} \frac{2 d t}{1 + t^{2}}$ $= 2 \int_{0}^{\infty} \frac{d t}{(1 + t^{2}) {(\frac{2 + 2 t^{2} + 1 - t^{2}}{1 + t^{2}})}^{2}}$ $= 2 \int_{0}^{\infty} \frac{{(1 + t^{2})}^{2}}{(1 + t^{2}) {(3 + t^{2})}^{2}} d t$ $= 2 \int_{0}^{\infty} \frac{t^{2} + 1}{{(t^{2} + 3)}^{2}} d t = \int_{- \infty}^{+ \infty} \frac{t^{2} + 1}{{(t^{2} + 3)}^{2}} d t l e t$ $c o n s i d e r t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z^{2} + 1}{{(z^{2} + 3)}^{2}}$ $φ (z) = \frac{z^{2} + 1}{{(z - i \sqrt{3})}^{2} {(z + i \sqrt{3})}^{2}} t h e p o l e s o f φ a r e$ $i \sqrt{3} a n d - i \sqrt{3} (d o u b l e s) R e s i d u s t h e o r e m g i v e$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i \sqrt{3})$ $R e s (φ, i \sqrt{3}) = {l i m}_{z \to i \sqrt{3}} \frac{1}{(2 - 1)!} {{(z - i \sqrt{3})}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i \sqrt{3}} {\frac{z^{2} + 1}{{(z + i \sqrt{3})}^{2}}}^{(1)}$ ${l i m}_{z \to i \sqrt{3}} {\frac{2 z {(z + i \sqrt{3})}^{2} - (z^{2} + 1) 2 (z + i \sqrt{3})}{{(z + i \sqrt{3})}^{4}}}$ $= {l i m}_{z \to i \sqrt{3}} {\frac{2 z (z + i \sqrt{3}) - 2 (z^{2} + 1)}{{(z + i \sqrt{3})}^{3}}}$ $= {l i m}_{z \to i \sqrt{3}} {\frac{2 z^{2} + 2 i z \sqrt{3} - 2 z^{2} - 2}{{(z + i \sqrt{3})}^{3}}}$ $= \frac{2 i (i \sqrt{3}) \sqrt{3} - 2}{{(2 i \sqrt{3})}^{3}} = \frac{- 8}{- 8 i 3 \sqrt{3}} = \frac{1}{i 3 \sqrt{3}}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{1}{i 3 \sqrt{3}} = \frac{2 π}{3 \sqrt{3}} = I_{1}$
Commented by abdo.msup.com last updated on 31/May/18
$I_2 =∫_π ^(2π) (dθ/((2+cosθ)^2 )) =_(θ =π+t) ∫_0 ^π (dt/((2−cost)^2 )) =_(tan((t/2))=x) ∫_0 ^(+∞) (1/((2−((1−x^2 )/(1+x^2 )))^2 )) ((2dx)/(1+x^2 )) = 2∫_0 ^(+∞) (dx/((1+x^2 ){((2+2x^2 −1+x^2 )/(1+x^2 ))}^2 )) = ∫_(−∞) ^(+∞) ((1+x^2 )/((3x^2 +1)^2 ))dx let consider the complex function ϕ(z) = ((z^2 +1)/((3z^2 +1)^2 )) ϕ(z) = ((z^2 +1)/(9( z −(i/(√3)))(z+(i/(√3))))) the poles of ϕ are (i/(√3)) and ((−i)/(√3))(double) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,(i/(√3))) Res(ϕ,(i/(√3)))=lim_(z→(i/(√3))) (1/((2−1)!)){ (z−(i/(√3)))ϕ(z)}^((1)) ...be continued...$
$I_{2} = \int_{π}^{2 π} \frac{d θ}{{(2 + c o s θ)}^{2}}$ $=_{θ = π + t} \int_{0}^{π} \frac{d t}{{(2 - c o s t)}^{2}}$ $=_{t a n (\frac{t}{2}) = x} \int_{0}^{+ \infty} \frac{1}{{(2 - \frac{1 - x^{2}}{1 + x^{2}})}^{2}} \frac{2 d x}{1 + x^{2}}$ $= 2 \int_{0}^{+ \infty} \frac{d x}{(1 + x^{2}) {\frac{2 + 2 x^{2} - 1 + x^{2}}{1 + x^{2}}}^{2}}$ $= \int_{- \infty}^{+ \infty} \frac{1 + x^{2}}{{(3 x^{2} + 1)}^{2}} d x l e t c o n s i d e r$ $t h e c o m p l e x f u n c t i o n$ $φ (z) = \frac{z^{2} + 1}{{(3 z^{2} + 1)}^{2}}$ $φ (z) = \frac{z^{2} + 1}{9 (z - \frac{i}{\sqrt{3}}) (z + \frac{i}{\sqrt{3}})} t h e p o l e s o f$ $φ a r e \frac{i}{\sqrt{3}} a n d \frac{- i}{\sqrt{3}} (d o u b l e)$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, \frac{i}{\sqrt{3}})$ $R e s (φ, \frac{i}{\sqrt{3}}) = {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{(2 - 1)!} {(z - \frac{i}{\sqrt{3}}) φ (z)}^{(1)}$ $. . . b e c o n t i n u e d . . .$
Commented by prof Abdo imad last updated on 01/Jun/18
$Res(ϕ,(i/(√3))) =lim_(z→(i/(√3))) { ((z^2 +1)/(9(z +(i/(√3)))^2 ))}^((1)) =(1/9){ ((2z(z+(i/(√3)))^2 −(z^2 +1)2(z +(i/(√3))))/((z +(i/(√3)))^4 ))} =lim_(z→(i/(√3))) (1/9){ ((2z(z +(i/(√3))) −2(z^2 +1))/((z +(i/(√3)))^3 ))} =lim_(z→(i/(√3))) (1/9){ (((2/(√3))iz −2)/((z +(i/(√3)))^3 ))} = (2/9) ((((i/(√3))(i/(√3)) −1)/((((2i)/(√3)))^3 )))= (2/9) (4/(3 (8/(3(√3)))i)) = ((√3)/(9i)) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((√3)/(9i)) = ((2π(√3))/9) =I_2 I = I_1 +I_2$
$R e s (φ, \frac{i}{\sqrt{3}}) = {l i m}_{z \to \frac{i}{\sqrt{3}}} {\frac{z^{2} + 1}{9 {(z + \frac{i}{\sqrt{3}})}^{2}}}^{(1)}$ $= \frac{1}{9} {\frac{2 z {(z + \frac{i}{\sqrt{3}})}^{2} - (z^{2} + 1) 2 (z + \frac{i}{\sqrt{3}})}{{(z + \frac{i}{\sqrt{3}})}^{4}}}$ $= {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{9} {\frac{2 z (z + \frac{i}{\sqrt{3}}) - 2 (z^{2} + 1)}{{(z + \frac{i}{\sqrt{3}})}^{3}}}$ $= {l i m}_{z \to \frac{i}{\sqrt{3}}} \frac{1}{9} {\frac{\frac{2}{\sqrt{3}} i z - 2}{{(z + \frac{i}{\sqrt{3}})}^{3}}}$ $= \frac{2}{9} (\frac{\frac{i}{\sqrt{3}} \frac{i}{\sqrt{3}} - 1}{{(\frac{2 i}{\sqrt{3}})}^{3}}) = \frac{2}{9} \frac{4}{3 \frac{8}{3 \sqrt{3}} i} = \frac{\sqrt{3}}{9 i}$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π \frac{\sqrt{3}}{9 i} = \frac{2 π \sqrt{3}}{9} = I_{2}$ $I = I_{1} + I_{2}$
Commented by prof Abdo imad last updated on 01/Jun/18

$I = \frac{2 π}{3 \sqrt{3}} + \frac{2 π \sqrt{3}}{9} = \frac{2 π \sqrt{3}}{9} + \frac{2 π \sqrt{3}}{9} \Rightarrow$ $I = \frac{4 π \sqrt{3}}{9} .$
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