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

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

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{d}\theta}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} } \\ $$

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

$${letut}\:\:{I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} } \\ $$$${I}\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{2}\:+{cos}\theta\right)^{\mathrm{2}} }\:\:+\:\int_{\pi} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\theta}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} }\:={I}_{\mathrm{1}} \:+\:{I}_{\mathrm{2}} \\ $$$${changement}\:{tan}\left(\frac{\theta}{\mathrm{2}}\right)={t}\:{give} \\ $$$${I}_{\mathrm{1}} \:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}\:+\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\frac{\mathrm{2}\:+\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\right)^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\:\mathrm{3}+{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{\left(\:{t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dt}\:=\int_{−\infty} ^{+\infty} \:\:\frac{{t}^{\mathrm{2}} \:+\mathrm{1}}{\left({t}^{\mathrm{2}} \:+\mathrm{3}\right)^{\mathrm{2}} }{dt}\:{let} \\ $$$${consider}\:{the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \left({z}\:+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\:{the}\:{poles}\:{of}\:\varphi\:{are} \\ $$$${i}\sqrt{\mathrm{3}}\:\:{and}\:−{i}\sqrt{\mathrm{3}}\left({doubles}\right)\:{Residus}\:{theorem}\:{give} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right) \\ $$$${Res}\left(\varphi,{i}\sqrt{\mathrm{3}}\right)\:={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\left\{\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left({z}\:+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$${lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}\:} \:\left\{\:\frac{\mathrm{2}{z}\left({z}\:+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\mathrm{2}\left({z}+{i}\sqrt{\mathrm{3}}\right)}{\left({z}\:+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\left\{\:\:\frac{\mathrm{2}{z}\left({z}\:+{i}\sqrt{\mathrm{3}}\right)\:−\mathrm{2}\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}{\left({z}\:+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\right\} \\ $$$$={lim}_{{z}\rightarrow{i}\sqrt{\mathrm{3}}} \:\:\left\{\:\frac{\mathrm{2}{z}^{\mathrm{2}} \:\:+\:\mathrm{2}{iz}\:\sqrt{\mathrm{3}}\:−\mathrm{2}{z}^{\mathrm{2}} \:−\mathrm{2}}{\left({z}+{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\right\} \\ $$$$=\:\frac{\mathrm{2}{i}\left({i}\sqrt{\mathrm{3}}\right)\sqrt{\mathrm{3}}\:−\mathrm{2}}{\left(\mathrm{2}{i}\sqrt{\mathrm{3}}\right)^{\mathrm{3}} }\:=\:\frac{−\mathrm{8}}{−\mathrm{8}{i}\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{1}}{{i}\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{{i}\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:{I}_{\mathrm{1}} \\ $$$$ \\ $$

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...

$${I}_{\mathrm{2}} =\int_{\pi} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\left(\mathrm{2}+{cos}\theta\right)^{\mathrm{2}} } \\ $$$$=_{\theta\:=\pi+{t}} \:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dt}}{\left(\mathrm{2}−{cost}\right)^{\mathrm{2}} } \\ $$$$=_{{tan}\left(\frac{{t}}{\mathrm{2}}\right)={x}} \:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\left(\mathrm{2}−\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\frac{\mathrm{2}{dx}}{\mathrm{1}+{x}^{\mathrm{2}} } \\ $$$$=\:\mathrm{2}\int_{\mathrm{0}} ^{+\infty} \:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left\{\frac{\mathrm{2}+\mathrm{2}{x}^{\mathrm{2}} \:−\mathrm{1}+{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\right\}^{\mathrm{2}} } \\ $$$$=\:\int_{−\infty} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{2}} }{\left(\mathrm{3}{x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} }{dx}\:\:{let}\:{consider} \\ $$$${the}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\left(\mathrm{3}{z}^{\mathrm{2}} \:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\varphi\left({z}\right)\:=\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{9}\left(\:{z}\:−\frac{{i}}{\sqrt{\mathrm{3}}}\right)\left({z}+\frac{{i}}{\sqrt{\mathrm{3}}}\right)}\:{the}\:{poles}\:{of} \\ $$$$\varphi\:{are}\:\frac{{i}}{\sqrt{\mathrm{3}}}\:\:{and}\:\frac{−{i}}{\sqrt{\mathrm{3}}}\left({double}\right) \\ $$$$\int_{−\infty} ^{+\infty} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\frac{{i}}{\sqrt{\mathrm{3}}}\right) \\ $$$${Res}\left(\varphi,\frac{{i}}{\sqrt{\mathrm{3}}}\right)={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{3}}}} \:\frac{\mathrm{1}}{\left(\mathrm{2}−\mathrm{1}\right)!}\left\{\:\left({z}−\frac{{i}}{\sqrt{\mathrm{3}}}\right)\varphi\left({z}\right)\right\}^{\left(\mathrm{1}\right)} \\ $$$$...{be}\:{continued}... \\ $$

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

$${Res}\left(\varphi,\frac{{i}}{\sqrt{\mathrm{3}}}\right)\:={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{3}}}} \:\:\left\{\:\:\frac{{z}^{\mathrm{2}} \:+\mathrm{1}}{\mathrm{9}\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }\right\}^{\left(\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{9}}\left\{\:\frac{\mathrm{2}{z}\left({z}+\frac{{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} \:−\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\mathrm{2}\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)}{\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{4}} }\right\} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{3}}}} \frac{\mathrm{1}}{\mathrm{9}}\left\{\:\frac{\mathrm{2}{z}\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)\:−\mathrm{2}\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)}{\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} }\right\} \\ $$$$={lim}_{{z}\rightarrow\frac{{i}}{\sqrt{\mathrm{3}}}} \frac{\mathrm{1}}{\mathrm{9}}\left\{\:\frac{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}{iz}\:−\mathrm{2}}{\left({z}\:+\frac{{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} }\right\} \\ $$$$=\:\frac{\mathrm{2}}{\mathrm{9}}\:\left(\frac{\frac{{i}}{\sqrt{\mathrm{3}}}\frac{{i}}{\sqrt{\mathrm{3}}}\:−\mathrm{1}}{\left(\frac{\mathrm{2}{i}}{\sqrt{\mathrm{3}}}\right)^{\mathrm{3}} }\right)=\:\frac{\mathrm{2}}{\mathrm{9}}\:\:\frac{\mathrm{4}}{\mathrm{3}\:\frac{\mathrm{8}}{\mathrm{3}\sqrt{\mathrm{3}}}{i}}\:=\:\:\frac{\sqrt{\mathrm{3}}}{\mathrm{9}{i}} \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\sqrt{\mathrm{3}}}{\mathrm{9}{i}}\:=\:\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:={I}_{\mathrm{2}} \\ $$$${I}\:=\:{I}_{\mathrm{1}} \:+{I}_{\mathrm{2}} \: \\ $$

Commented by prof Abdo imad last updated on 01/Jun/18

I = ((2π)/(3(√3))) + ((2π(√3))/9) = ((2π(√3))/9) +((2π(√3))/9) ⇒ I = ((4π(√3))/9) .

$${I}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:\:+\:\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:=\:\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:+\frac{\mathrm{2}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:\Rightarrow \\ $$$${I}\:=\:\frac{\mathrm{4}\pi\sqrt{\mathrm{3}}}{\mathrm{9}}\:. \\ $$

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