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Question Number 36197 by prof Abdo imad last updated on 30/May/18

find the value of ∫_0 ^(2π) (dx/(cos^2 x +3 sin^2 x))

$${find}\:{the}\:{value}\:{of}\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dx}}{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}\:{sin}^{\mathrm{2}} {x}} \\ $$

Commented by maxmathsup by imad last updated on 14/Aug/18

let A = ∫_0 ^(2π) (dx/(cos^2 x +3sin^2 x)) A = ∫_0 ^(2π) (dx/(((1+cos(2x))/2)+((1−sin(2x))/2))) = ∫_0 ^(2π) ((2dx)/(2 +cos(2x)−sin(2x))) =_(2x=t) ∫_0 ^(4π) (dt/(2+cost −sint)) = ∫_0 ^(2π) (dt/(2+cost −sint)) + ∫_(2π) ^(4π) (dt/(2+cost−sint)) but ∫_(2π) ^(4π) (dt/(2+cost −sint)) =_(t=2π +α) ∫_0 ^(2π) (dα/(2+cosα−sinα)) ⇒ A = ∫_0 ^(2π) ((2dt)/(2+cost −sint)) changement e^(it) =z give A = ∫_(∣z∣=1) (2/(2 +((z+z^(−1) )/2) +((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((4dz)/(iz(4 +z +z^(−1) −i(z−z^− )))) = ∫_(∣z∣=1) ((−4idz)/(4z +z^2 +1 −iz^2 +i)) = ∫_(∣z∣=1) ((−4iz)/((1−i)z^2 +4z +1+i)) let consider the compolex function ϕ(z) = ((−4i)/((1−i)z^2 +4z +1+i)) poles of ϕ? Δ^′ =4 −(1−i)(1+i) =4−2=2 ⇒ z_1 =((−2 +(√2))/(1−i)) =(((√2)(1−(√2)))/((√2)e^(−((iπ)/4)) )) =(1−(√2))e^((iπ)/4) z_2 = ((−2−(√2))/(1−i)) =(((√2)(−1−(√2)))/((√2)e^(−((iπ)/4)) )) =(−1−(√2)) e^((iπ)/4) ∣z_1 ∣ =(√2)−1<1 and ∣z_2 ∣ =(√2)+1 >1 (to eliminate from residus) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) but ϕ(z) =((−4i)/((1−i)(z−z_1 )(z−z_2 ))) Res(ϕ,z_1 ) =((−4i)/((1−i)(z_1 −z_2 ))) = ((−4i)/((1−i)2 e^(i(π/4)) )) = ((−2i)/(√2)) =((−i)/(√2)) ⇒ ∫_(∣z∣=1 ) ϕ(z)dz =2iπ(((−i)/(√2))) =((2π)/(√2)) ⇒ A = ((2π)/(√2)) ⇒ A =π(√2).

$${let}\:\:{A}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{{cos}^{\mathrm{2}} {x}\:+\mathrm{3}{sin}^{\mathrm{2}} {x}} \\ $$$${A}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\frac{\mathrm{1}−{sin}\left(\mathrm{2}{x}\right)}{\mathrm{2}}}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}\:\:+{cos}\left(\mathrm{2}{x}\right)−{sin}\left(\mathrm{2}{x}\right)} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+{cost}\:−{sint}}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dt}}{\mathrm{2}+{cost}\:−{sint}}\:+\:\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+{cost}−{sint}} \\ $$$${but}\:\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+{cost}\:−{sint}}\:=_{{t}=\mathrm{2}\pi\:+\alpha} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{d}\alpha}{\mathrm{2}+{cos}\alpha−{sin}\alpha}\:\Rightarrow \\ $$$${A}\:=\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{cost}\:−{sint}}\:\:{changement}\:{e}^{{it}} \:={z}\:{give} \\ $$$${A}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}\:+\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{4}{dz}}{{iz}\left(\mathrm{4}\:+{z}\:+{z}^{−\mathrm{1}} \:−{i}\left({z}−{z}^{−} \right)\right)}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{−\mathrm{4}{idz}}{\mathrm{4}{z}\:+{z}^{\mathrm{2}} \:+\mathrm{1}\:−{iz}^{\mathrm{2}} \:+{i}} \\ $$$$=\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{−\mathrm{4}{iz}}{\left(\mathrm{1}−{i}\right){z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}+{i}}\:\:{let}\:{consider}\:{the}\:{compolex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\frac{−\mathrm{4}{i}}{\left(\mathrm{1}−{i}\right){z}^{\mathrm{2}} \:+\mathrm{4}{z}\:+\mathrm{1}+{i}}\:\:{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\mathrm{4}\:−\left(\mathrm{1}−{i}\right)\left(\mathrm{1}+{i}\right)\:=\mathrm{4}−\mathrm{2}=\mathrm{2}\:\Rightarrow\:{z}_{\mathrm{1}} =\frac{−\mathrm{2}\:+\sqrt{\mathrm{2}}}{\mathrm{1}−{i}}\:=\frac{\sqrt{\mathrm{2}}\left(\mathrm{1}−\sqrt{\mathrm{2}}\right)}{\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=\left(\mathrm{1}−\sqrt{\mathrm{2}}\right){e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$${z}_{\mathrm{2}} =\:\frac{−\mathrm{2}−\sqrt{\mathrm{2}}}{\mathrm{1}−{i}}\:=\frac{\sqrt{\mathrm{2}}\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)}{\sqrt{\mathrm{2}}{e}^{−\frac{{i}\pi}{\mathrm{4}}} }\:=\left(−\mathrm{1}−\sqrt{\mathrm{2}}\right)\:{e}^{\frac{{i}\pi}{\mathrm{4}}} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:=\sqrt{\mathrm{2}}−\mathrm{1}<\mathrm{1}\:\:{and}\:\:\mid{z}_{\mathrm{2}} \mid\:=\sqrt{\mathrm{2}}+\mathrm{1}\:>\mathrm{1}\:\left({to}\:{eliminate}\:{from}\:{residus}\right)\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:\:{but}\:\:\varphi\left({z}\right)\:=\frac{−\mathrm{4}{i}}{\left(\mathrm{1}−{i}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\frac{−\mathrm{4}{i}}{\left(\mathrm{1}−{i}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\:\frac{−\mathrm{4}{i}}{\left(\mathrm{1}−{i}\right)\mathrm{2}\:{e}^{{i}\frac{\pi}{\mathrm{4}}} }\:=\:\frac{−\mathrm{2}{i}}{\sqrt{\mathrm{2}}}\:=\frac{−{i}}{\sqrt{\mathrm{2}}}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}\:} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left(\frac{−{i}}{\sqrt{\mathrm{2}}}\right)\:=\frac{\mathrm{2}\pi}{\sqrt{\mathrm{2}}}\:\Rightarrow\:{A}\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{2}}}\:\:\Rightarrow\:{A}\:=\pi\sqrt{\mathrm{2}}. \\ $$

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