solve-the-integral-with-Residue-theorem-0-2pi-3-d-9-sin-2-

Question Number 75435 by Crabby89p13 last updated on 11/Dec/19

solve the integral with Residue theorem. ∫_0 ^(2π) ((3 dθ)/(9 +sin^2 θ))

$${solve}\:{the}\:{integral}\:{with}\:{Residue}\:{theorem}. \\ $$$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{\mathrm{3}\:{d}\theta}{\mathrm{9}\:+\mathrm{sin}^{\mathrm{2}} \theta} \\ $$

Commented by mathmax by abdo last updated on 11/Dec/19

I =∫_0 ^(2π) ((3dθ)/(9+sin^2 θ)) ⇒I =∫_0 ^(2π) ((3dθ)/(9+((1−cos(2θ))/2))) =∫_0 ^(2π) ((6dθ)/(18+1−cos(2θ))) =_(2θ =t) ∫_0 ^(4π) (6/(19−cost))(dt/2) =3 ∫_0 ^(4π) (dt/(19−cost)) =3{ ∫_0 ^(2π) (dt/(19−cost)) +∫_(2π) ^(4π) (dt/(19−cost))} but ∫_(2π) ^(4π) (dt/(19−cost)) =_(t=2π +u) ∫_0 ^(2π) (du/(19−cosu)) ⇒I =6 ∫_0 ^(2π) (dt/(19−cost)) we have ∫_0 ^(2π) (dt/(19−cost)) =_(e^(it) =z) ∫_(∣z∣=1) (1/(19−((z+z^(−1) )/2)))(dz/(iz)) =∫_(∣z∣=1) ((2dz)/(iz{38−z−z^(−1) })) =∫_(∣z∣=1) ((−2idz)/(38z−z^2 −1)) =∫_(∣z∣=1) ((2idz)/(z^2 −38z +1)) let W(z)=((2i)/(z^2 −38z +1)) poles of W? z^2 −38z +1=0→Δ^′ =19^2 −1 =360 z_1 =19+6(√(10)) and z_2 =19−6(√(10)) ∣z_1 ∣−1 =18+6(√(10))>0 (z_1 is out of circle) ∣z_2 ∣−1 =18−6(√(10)) <0 ⇒ ∫_(∣z∣=1) W(z)dz =2iπ Res(W,z_2 ) =2iπ×((2i)/((z_2 −z_1 ))) =((−4π)/(−12(√(10)))) =(π/(3(√(10)))) ⇒ I =6×(π/(3(√(10)))) =((2π)/( (√(10))))

$${I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{3}{d}\theta}{\mathrm{9}+{sin}^{\mathrm{2}} \theta}\:\Rightarrow{I}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{3}{d}\theta}{\mathrm{9}+\frac{\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)}{\mathrm{2}}}\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{6}{d}\theta}{\mathrm{18}+\mathrm{1}−{cos}\left(\mathrm{2}\theta\right)} \\ $$$$=_{\mathrm{2}\theta\:={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\frac{\mathrm{6}}{\mathrm{19}−{cost}}\frac{{dt}}{\mathrm{2}}\:=\mathrm{3}\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\frac{{dt}}{\mathrm{19}−{cost}} \\ $$$$=\mathrm{3}\left\{\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{\mathrm{19}−{cost}}\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\frac{{dt}}{\mathrm{19}−{cost}}\right\}\:{but} \\ $$$$\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\frac{{dt}}{\mathrm{19}−{cost}}\:=_{{t}=\mathrm{2}\pi\:+{u}} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{du}}{\mathrm{19}−{cosu}}\:\Rightarrow{I}\:=\mathrm{6}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{\mathrm{19}−{cost}} \\ $$$${we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{dt}}{\mathrm{19}−{cost}}\:=_{{e}^{{it}} ={z}} \:\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{\mathrm{19}−\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\frac{{dz}}{{iz}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{dz}}{{iz}\left\{\mathrm{38}−{z}−{z}^{−\mathrm{1}} \right\}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{−\mathrm{2}{idz}}{\mathrm{38}{z}−{z}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{2}{idz}}{{z}^{\mathrm{2}} −\mathrm{38}{z}\:+\mathrm{1}}\:\:{let}\:{W}\left({z}\right)=\frac{\mathrm{2}{i}}{{z}^{\mathrm{2}} −\mathrm{38}{z}\:+\mathrm{1}}\:\:{poles}\:{of}\:{W}? \\ $$$${z}^{\mathrm{2}} −\mathrm{38}{z}\:+\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{19}^{\mathrm{2}} −\mathrm{1}\:=\mathrm{360} \\ $$$${z}_{\mathrm{1}} =\mathrm{19}+\mathrm{6}\sqrt{\mathrm{10}}\:\:\:\:{and}\:{z}_{\mathrm{2}} =\mathrm{19}−\mathrm{6}\sqrt{\mathrm{10}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\mathrm{18}+\mathrm{6}\sqrt{\mathrm{10}}>\mathrm{0}\:\:\:\:\left({z}_{\mathrm{1}} {is}\:{out}\:{of}\:{circle}\right) \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\mathrm{18}−\mathrm{6}\sqrt{\mathrm{10}}\:\:<\mathrm{0}\:\:\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {W}\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left({W},{z}_{\mathrm{2}} \right)\:=\mathrm{2}{i}\pi×\frac{\mathrm{2}{i}}{\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\frac{−\mathrm{4}\pi}{−\mathrm{12}\sqrt{\mathrm{10}}}\:=\frac{\pi}{\mathrm{3}\sqrt{\mathrm{10}}}\:\Rightarrow\:{I}\:=\mathrm{6}×\frac{\pi}{\mathrm{3}\sqrt{\mathrm{10}}}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{10}}} \\ $$$$ \\ $$

Commented by Crabby89p13 last updated on 11/Dec/19