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Question Number 37587 by prof Abdo imad last updated on 15/Jun/18

calculate ∫_0 ^(2π) (dx/(2cos^2 x +(√3) sin^2 x))

$${calculate}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{2}{cos}^{\mathrm{2}} {x}\:+\sqrt{\mathrm{3}}\:{sin}^{\mathrm{2}} {x}} \\ $$

Commented by math khazana by abdo last updated on 16/Jun/18

I = ∫_0 ^(2π) (dx/(2 ((1+cos(2x))/2)+(√3)((1−cos(2x))/2))) = ∫_0 ^(2π) (dx/(1+cos(2x)+((√3)/2)(1−cos(2x)))) = ∫_0 ^(2π) ((2dx)/(2 +2cos(2x) +(√3) −(√3)cos(2x))) = ∫_0 ^(2π) ((2dx)/(2+(√3) +(2−(√3))cos(2x))) =_(2x=t) ∫_0 ^(4π) ((2dx)/(2+(√3) +(2−(√3))cost)) (dt/2) = ∫_0 ^(4π) (dt/(2 +(√3) +(2−(√3))cost)) = ∫_0 ^(2π) (dt/(2+(√3) +(2−(√3))cost)) +∫_(2π) ^(4π) (dt/(2+(√3) +(2−(√3))cost)) =K +H chang. e^(it) =z give K = ∫_(∣z∣ =1) (1/(2+(√3) +(2−(√3))((z+z^(−1) )/2))) (dz/(iz)) = ∫_(∣z∣=1) ((2dz)/(iz(4+2(√3) +(2−(√3))(z+z^(−1) ))) = ∫_(∣z∣=1) ((−2idz)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) let ϕ(z)= ((−2i)/((2−(√3))z^2 +(4+2(√3))z +2−(√3))) poles of ϕ? Δ^′ = (2+(√3))^2 −(2−(√3))^2 =4 +4(√3) +3−4 +4(√3) −3 =8(√3) z_1 = ((−2−(√3) +2(√2)(√(√3)))/(2−(√3))) z_2 = ((−2−(√3) −2(√2) (√(√3)))/(2−(√3))) ∣z_1 ∣ −1 =((∣ 2+(√3) −2(√2)(√(√3))∣)/(2−(√3))) (2+(√3))^2 −8(√3) = 7 −4(√3) −8(√3)=7−12(√3)<0 ⇒∣z_1 ∣<1 and we verify that ∣z_2 ∣>1 ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) Res(ϕ,z_1 ) = ((−2i)/((2−(√3))(z_1 −z_2 ))) = ((−2i)/((2−(√3))((4(√(2(√3))))/(2−(√3))))) = ((−i)/(2(√(2(√3))))) .⇒ ∫_(∣z∣=1) ϕ(z)dz = 2iπ ((−i)/(2(√(2(√3))))) = (π/(√(2(√3)))) =K .

$${I}\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\mathrm{2}\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\sqrt{\mathrm{3}}\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\left(\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}\:+\mathrm{2}{cos}\left(\mathrm{2}{x}\right)\:+\sqrt{\mathrm{3}}\:−\sqrt{\mathrm{3}}{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cos}\left(\mathrm{2}{x}\right)} \\ $$$$=_{\mathrm{2}{x}={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}\:+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$={K}\:+{H} \\ $$$${chang}.\:{e}^{{it}} \:={z}\:{give} \\ $$$${K}\:=\:\int_{\mid{z}\mid\:=\mathrm{1}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\:\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}+{z}^{−\mathrm{1}} \right)\right.} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{−\mathrm{2}{idz}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} \:+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${let}\:\varphi\left({z}\right)=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){z}^{\mathrm{2}} \:+\left(\mathrm{4}+\mathrm{2}\sqrt{\mathrm{3}}\right){z}\:+\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\:\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)^{\mathrm{2}} =\mathrm{4}\:+\mathrm{4}\sqrt{\mathrm{3}}\:+\mathrm{3}−\mathrm{4}\:+\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{3} \\ $$$$=\mathrm{8}\sqrt{\mathrm{3}} \\ $$$${z}_{\mathrm{1}} =\:\frac{−\mathrm{2}−\sqrt{\mathrm{3}}\:\:\:+\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$${z}_{\mathrm{2}} =\:\frac{−\mathrm{2}−\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}\:\sqrt{\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\frac{\mid\:\mathrm{2}+\sqrt{\mathrm{3}}\:−\mathrm{2}\sqrt{\mathrm{2}}\sqrt{\sqrt{\mathrm{3}}}\mid}{\mathrm{2}−\sqrt{\mathrm{3}}} \\ $$$$\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{\mathrm{2}} \:−\mathrm{8}\sqrt{\mathrm{3}}\:=\:\mathrm{7}\:−\mathrm{4}\sqrt{\mathrm{3}}\:−\mathrm{8}\sqrt{\mathrm{3}}=\mathrm{7}−\mathrm{12}\sqrt{\mathrm{3}}<\mathrm{0} \\ $$$$\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1}\:{and}\:{we}\:{verify}\:{that}\:\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\left({z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} \right)} \\ $$$$=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\frac{\mathrm{4}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}{\mathrm{2}−\sqrt{\mathrm{3}}}}\:=\:\:\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:.\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\varphi\left({z}\right){dz}\:=\:\mathrm{2}{i}\pi\:\frac{−{i}}{\mathrm{2}\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:=\:\frac{\pi}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\:={K}\:. \\ $$

Commented by math khazana by abdo last updated on 16/Jun/18

H =∫_(2π) ^(4π) (dt/(2+(√3) +(2−(√3))cost)) =_(t =2π +x) ∫_0 ^(2π) (dx/(2+(√(3 ))+(2−(√3))cosx)) =K =(π/(√(2(√3)))) ⇒ I =H +K = ((2π)/(√(2(√3)))) = ((π(√2))/(√(√3))) I =((π(√2))/(√(√3))) .

$${H}\:=\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\:\frac{{dt}}{\mathrm{2}+\sqrt{\mathrm{3}}\:+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cost}} \\ $$$$=_{{t}\:=\mathrm{2}\pi\:+{x}} \:\:\:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\:\frac{{dx}}{\mathrm{2}+\sqrt{\mathrm{3}\:}+\left(\mathrm{2}−\sqrt{\mathrm{3}}\right){cosx}}\:\:\:={K} \\ $$$$=\frac{\pi}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:\Rightarrow \\ $$$${I}\:={H}\:+{K}\:=\:\frac{\mathrm{2}\pi}{\sqrt{\mathrm{2}\sqrt{\mathrm{3}}}}\:=\:\frac{\pi\sqrt{\mathrm{2}}}{\sqrt{\sqrt{\mathrm{3}}}}\:\: \\ $$$${I}\:\:=\frac{\pi\sqrt{\mathrm{2}}}{\sqrt{\sqrt{\mathrm{3}}}}\:. \\ $$

Answered by ajfour last updated on 15/Jun/18

∫_0 ^( 2π) ((sec^2 xdx)/(2+(√3)tan^2 x)) =(4/(√3)) ∫_0 ^( ∞) (dt/(t^2 +(2/(√3)))) = (4/(√3))×(3^(1/4) /(√2))((π/2)) =π (√(2/(√3))) .

$$\int_{\mathrm{0}} ^{\:\:\mathrm{2}\pi} \frac{\mathrm{sec}\:^{\mathrm{2}} {xdx}}{\mathrm{2}+\sqrt{\mathrm{3}}\mathrm{tan}\:^{\mathrm{2}} {x}} \\ $$$$=\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}\:\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dt}}{{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}} \\ $$$$=\:\frac{\mathrm{4}}{\sqrt{\mathrm{3}}}×\frac{\mathrm{3}^{\mathrm{1}/\mathrm{4}} }{\sqrt{\mathrm{2}}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\pi\:\sqrt{\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}}\:\:. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jun/18

Commented by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18

in place of (Π/2) i put (Π/4)

$${in}\:{place}\:{of}\:\frac{\Pi}{\mathrm{2}}\:\:\:\:{i}\:{put}\:\frac{\Pi}{\mathrm{4}}\:\:\: \\ $$

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