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Question Number 64649 by mathmax by abdo last updated on 20/Jul/19

calculate ∫_0 ^(2π) ((cosθ)/(5+3cosθ))dθ

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{cos}\theta}{\mathrm{5}+\mathrm{3}{cos}\theta}{d}\theta \\ $$

Commented by mathmax by abdo last updated on 20/Jul/19

let A =∫_0 ^(2π) ((cosθ)/(5+3cosθ)) ⇒A =(1/3) ∫_0 ^(2π) ((3cosθ +5−5)/(3cosθ +5))dθ =((2π)/3) −(5/3) ∫_0 ^(2π) (dθ/(3cosθ +5)) changement e^(iθ) =z give ∫_0 ^(2π) (dθ/(3cosθ +5)) =∫_(∣z∣=1) (1/(3 ((z+z^(−1) )/2)+5)) (dz/(iz)) = ∫_(∣z∣=1) ((2dz)/(iz(3z+3z^(−1) +10))) =∫_(∣z∣=1) ((−2i)/(3z^2 +3 +10z))dz let W(z) =((−2i)/(3z^2 +10z +3)) poles of W? Δ^′ =5^2 −9 =16 ⇒z_1 =((−5 +4)/3) =−(1/3) and z_2 =((−5−4)/3) =−3 ⇒ W(z) =((−2i)/(3(z−z_1 )(z−z_2 ))) (z_2 is out of circle) residus theorem give ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_1 ) Res(W,z_1 ) =lim_(z→z_1 ) (z−z_1 )W(z) =((−2i)/(3(z_1 −z_2 ))) =((−2i)/(3(−(1/3)+3))) =((−2i)/(3.(8/3))) =−(i/4) ⇒∫_(∣z∣=1) W(z)dz =2iπ(−(i/4)) =(π/2) ⇒ A =((2π)/3) −(5/3) (π/2) =((2π)/3)−((5π)/6) =((4π−5π)/6) ⇒ A =−(π/6) .

$${let}\:{A}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}\theta}{\mathrm{5}+\mathrm{3}{cos}\theta}\:\Rightarrow{A}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{\mathrm{3}{cos}\theta\:+\mathrm{5}−\mathrm{5}}{\mathrm{3}{cos}\theta\:+\mathrm{5}}{d}\theta \\ $$$$=\frac{\mathrm{2}\pi}{\mathrm{3}}\:−\frac{\mathrm{5}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\mathrm{3}{cos}\theta\:+\mathrm{5}}\:\:{changement}\:{e}^{{i}\theta} ={z}\:{give} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\mathrm{3}{cos}\theta\:+\mathrm{5}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{3}\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}+\mathrm{5}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{3}{z}+\mathrm{3}{z}^{−\mathrm{1}} \:+\mathrm{10}\right)}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{2}{i}}{\mathrm{3}{z}^{\mathrm{2}} \:+\mathrm{3}\:+\mathrm{10}{z}}{dz}\:{let} \\ $$$${W}\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\mathrm{3}{z}^{\mathrm{2}} \:+\mathrm{10}{z}\:+\mathrm{3}}\:\:{poles}\:{of}\:{W}? \\ $$$$\Delta^{'} =\mathrm{5}^{\mathrm{2}} −\mathrm{9}\:=\mathrm{16}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{5}\:+\mathrm{4}}{\mathrm{3}}\:=−\frac{\mathrm{1}}{\mathrm{3}}\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{5}−\mathrm{4}}{\mathrm{3}}\:=−\mathrm{3}\:\Rightarrow \\ $$$${W}\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\mathrm{3}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\:\:\:\:\left({z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle}\right)\:{residus}\:{theorem}\:{give} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{z}_{\mathrm{1}} \right) \\ $$$${Res}\left({W},{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right){W}\left({z}\right)\:=\frac{−\mathrm{2}{i}}{\mathrm{3}\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\frac{−\mathrm{2}{i}}{\mathrm{3}\left(−\frac{\mathrm{1}}{\mathrm{3}}+\mathrm{3}\right)} \\ $$$$=\frac{−\mathrm{2}{i}}{\mathrm{3}.\frac{\mathrm{8}}{\mathrm{3}}}\:=−\frac{{i}}{\mathrm{4}}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(−\frac{{i}}{\mathrm{4}}\right)\:=\frac{\pi}{\mathrm{2}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{2}\pi}{\mathrm{3}}\:−\frac{\mathrm{5}}{\mathrm{3}}\:\frac{\pi}{\mathrm{2}}\:=\frac{\mathrm{2}\pi}{\mathrm{3}}−\frac{\mathrm{5}\pi}{\mathrm{6}}\:=\frac{\mathrm{4}\pi−\mathrm{5}\pi}{\mathrm{6}}\:\Rightarrow\:{A}\:=−\frac{\pi}{\mathrm{6}}\:. \\ $$$$ \\ $$

Answered by MJS last updated on 20/Jul/19

∫_0 ^(2π) ((cos θ)/(5+3cos θ))dθ=2∫_0 ^π ((cos θ)/(5+3cos θ))dθ= [t=tan (θ/2) → dθ=2cos^2 (θ/2) dt] −2∫_0 ^∞ ((t^2 −1)/((t^2 +1)(t^2 +4)))dt=(4/3)∫_0 ^∞ (dt/(t^2 +1))−((10)/3)∫_0 ^∞ (dt/(t^2 +4))= =[(4/3)arctan t −(5/3)arctan (t/2)]_0 ^∞ =−(π/6)

$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{cos}\:\theta}{\mathrm{5}+\mathrm{3cos}\:\theta}{d}\theta=\mathrm{2}\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{\mathrm{cos}\:\theta}{\mathrm{5}+\mathrm{3cos}\:\theta}{d}\theta= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\:\rightarrow\:{d}\theta=\mathrm{2cos}^{\mathrm{2}} \:\frac{\theta}{\mathrm{2}}\:{dt}\right] \\ $$$$−\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{t}^{\mathrm{2}} −\mathrm{1}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{4}\right)}{dt}=\frac{\mathrm{4}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{1}}−\frac{\mathrm{10}}{\mathrm{3}}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{{t}^{\mathrm{2}} +\mathrm{4}}= \\ $$$$=\left[\frac{\mathrm{4}}{\mathrm{3}}\mathrm{arctan}\:{t}\:−\frac{\mathrm{5}}{\mathrm{3}}\mathrm{arctan}\:\frac{{t}}{\mathrm{2}}\right]_{\mathrm{0}} ^{\infty} =−\frac{\pi}{\mathrm{6}} \\ $$

Commented by mathmax by abdo last updated on 20/Jul/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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