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Question Number 28882 by abdo imad last updated on 31/Jan/18
find ∫_(−π) ^π ((2dt)/(2+sint +cost)) .
$${find}\:\int_{−\pi} ^{\pi} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}+{sint}\:+{cost}}\:. \\ $$
Commented by abdo imad last updated on 02/Feb/18
let use the ch. e^(it) =z I= ∫_(∣z∣=1) (2/(2 +((z−z^− )/(2i)) +((z+z^(−1) )/2))) (dz/(iz)) I= ∫_(∣z∣=1) (1/(iz( 1+ ((z−z^(−1) )/(4i)) +((z+z^(−1) )/4))))dz I= ∫_(∣z∣=1) (dz/(z(i + ((z−z^(−1) )/4) +((i(z+z^(−1) ))/4)))) I= ∫_(∣z∣=1) ((4dz)/(z(4i +z−z^(−1) +iz +iz^(−1) ))) I= ∫_(∣z∣=1) ((4dz)/(4iz +z^2 −1 +iz^2 +i)) = ∫_(∣z∣=1) ((4dz)/((1+i)z^2 +4iz −1+i)) let introduce the complex function w(z)= (4/((1+i)z^2 +4iz−1+i)) poles of w? Δ^′ =(2i)^2 −(1+i)(−1+i)=−4 −(−2) =−2 =(i(√2))^2 z_1 =((−2i +i(√2))/(1+i)) and z_2 =((−2i−i(√2))/(1+i)) ∣z_1 ∣= ((∣−2i +i(√2)∣)/(∣1+i∣)) =((2−(√2))/( (√2)))=(√2) −1 and ∣z_1 ∣−1=(√2)−2<0 so ∣z_1 ∣<1
$${let}\:{use}\:{the}\:{ch}.\:{e}^{{it}} ={z} \\ $$$${I}=\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}\:+\frac{{z}−{z}^{−} }{\mathrm{2}{i}}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$${I}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{iz}\left(\:\mathrm{1}+\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{4}{i}}\:+\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{4}}\right)}{dz} \\ $$$${I}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{dz}}{{z}\left({i}\:+\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{4}}\:+\frac{{i}\left({z}+{z}^{−\mathrm{1}} \right)}{\mathrm{4}}\right)} \\ $$$${I}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{4}{dz}}{{z}\left(\mathrm{4}{i}\:+{z}−{z}^{−\mathrm{1}} \:+{iz}\:+{iz}^{−\mathrm{1}} \right)} \\ $$$${I}=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{4}{dz}}{\mathrm{4}{iz}\:+{z}^{\mathrm{2}} −\mathrm{1}\:+{iz}^{\mathrm{2}} +{i}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{4}{dz}}{\left(\mathrm{1}+{i}\right){z}^{\mathrm{2}} \:+\mathrm{4}{iz}\:−\mathrm{1}+{i}}\:{let}\:{introduce}\:{the}\:{complex} \\ $$$${function}\:{w}\left({z}\right)=\:\:\:\frac{\mathrm{4}}{\left(\mathrm{1}+{i}\right){z}^{\mathrm{2}} \:+\mathrm{4}{iz}−\mathrm{1}+{i}}\:\:\:{poles}\:{of}\:{w}? \\ $$$$\Delta^{'} =\left(\mathrm{2}{i}\right)^{\mathrm{2}} \:−\left(\mathrm{1}+{i}\right)\left(−\mathrm{1}+{i}\right)=−\mathrm{4}\:−\left(−\mathrm{2}\right)\:=−\mathrm{2}\:=\left({i}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} \\ $$$${z}_{\mathrm{1}} =\frac{−\mathrm{2}{i}\:+{i}\sqrt{\mathrm{2}}}{\mathrm{1}+{i}}\:\:\:{and}\:{z}_{\mathrm{2}} =\frac{−\mathrm{2}{i}−{i}\sqrt{\mathrm{2}}}{\mathrm{1}+{i}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\:\frac{\mid−\mathrm{2}{i}\:+{i}\sqrt{\mathrm{2}}\mid}{\mid\mathrm{1}+{i}\mid}\:=\frac{\mathrm{2}−\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}=\sqrt{\mathrm{2}}\:−\mathrm{1}\:{and}\:\mid{z}_{\mathrm{1}} \mid−\mathrm{1}=\sqrt{\mathrm{2}}−\mathrm{2}<\mathrm{0} \\ $$$${so}\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$
Commented by abdo imad last updated on 02/Feb/18
∣z_2 ∣= ((2+(√2))/( (√2))) =(√2)+1⇒∣z_2 ∣−1=(√2)>0( to eliminate frm residus) ∫_(∣z∣=1) w(z)dz=2iπRes(w,z_1 ) but w(z)= (4/((1+i)(z−z_1 )(z−z_2 )))⇒ Res(w,z_1 )= (4/((1+i)(z_1 −z_2 )))= (4/((1+i)((2i(√2))/(1+i)))) = (2/(i(√2))) =((√2)/i) ∫_(∣z∣=1) w(z)dz=2iπ.((√2)/i) = 2π(√2) .⇒ I=2π(√2) .
$$\mid{z}_{\mathrm{2}} \mid=\:\frac{\mathrm{2}+\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}+\mathrm{1}\Rightarrow\mid{z}_{\mathrm{2}} \mid−\mathrm{1}=\sqrt{\mathrm{2}}>\mathrm{0}\left(\:{to}\:{eliminate}\:{frm}\:{residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {w}\left({z}\right){dz}=\mathrm{2}{i}\pi{Res}\left({w},{z}_{\mathrm{1}} \right)\:{but}\: \\ $$$${w}\left({z}\right)=\:\:\:\:\:\frac{\mathrm{4}}{\left(\mathrm{1}+{i}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\Rightarrow \\ $$$${Res}\left({w},{z}_{\mathrm{1}} \right)=\:\:\frac{\mathrm{4}}{\left(\mathrm{1}+{i}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}=\:\:\:\frac{\mathrm{4}}{\left(\mathrm{1}+{i}\right)\frac{\mathrm{2}{i}\sqrt{\mathrm{2}}}{\mathrm{1}+{i}}}\:=\:\frac{\mathrm{2}}{{i}\sqrt{\mathrm{2}}}\:=\frac{\sqrt{\mathrm{2}}}{{i}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} {w}\left({z}\right){dz}=\mathrm{2}{i}\pi.\frac{\sqrt{\mathrm{2}}}{{i}}\:=\:\mathrm{2}\pi\sqrt{\mathrm{2}}\:.\Rightarrow\:\:{I}=\mathrm{2}\pi\sqrt{\mathrm{2}}\:. \\ $$$$ \\ $$

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