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Question Number 35225 by abdo mathsup 649 cc last updated on 16/May/18
1) find f(a) = ∫_0 ^(2π) (dt/(a cos^2 t + sin^2 t)) with a≠0 2) find g(a) = ∫_0 ^(2π) ((cos^2 t)/((a cos^2 t +sin^2 t)^2 ))dt
$$\left.\mathrm{1}\right)\:{find}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{a}\:{cos}^{\mathrm{2}} {t}\:+\:{sin}^{\mathrm{2}} {t}}\:{with}\:{a}\neq\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{g}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{cos}^{\mathrm{2}} {t}}{\left({a}\:{cos}^{\mathrm{2}} {t}\:+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\: \\ $$
Commented by abdo mathsup 649 cc last updated on 16/May/18
a>0
$${a}>\mathrm{0} \\ $$
Commented by prof Abdo imad last updated on 20/May/18
we have f(a) = ∫_0 ^(2π) (dt/(a((1+cos(2t))/2) +((1−cos(t))/2))) = 2 ∫_0 ^(2π) (dt/(a +a cos(2t) +1 −cos(2t))) =2 ∫_0 ^(2π) (dt/(a+1 +(a−1)cos(2t))) =_(2t=x) 2 ∫_0 ^(4π) (1/(a+1 +(a−1)cosx)) (dx/2) = ∫_0 ^(2π) (dx/(a+1 +(a−1)cosx)) +∫_(2π) ^(4π) (dx/(a+1 +(a−1)cosx)) ∫_(2π) ^(4π) (dx/(a+1 +(a−1)cosx)) =_(x=t+2π) ∫_0 ^(2π) (dt/(a+1 +(a−1)cost)) ⇒ f(a) = 2 ∫_0 ^(2π) (dx/(a+1 +(a−1)cosx)) changement e^(ix) =z give f(a) = 2∫_(∣z∣=1) (1/(a+1 +(a−1)((z+z^(−1) )/2))) (dz/(iz)) f(a) =4 ∫_(∣z∣=1) (dz/(iz(2+2a +(a−1)(z+z^(−1) )))) =∫_(∣z∣=1) ((−4i dz)/(2(1+a)z +(a−1)z^2 +a−1)) let inyroduce tbe complex function ϕ(z) = ((−4i)/((a−1)z^2 +2(a+1)z +a−1)) .poles of ϕ?
$${we}\:{have}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{a}\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\:+\frac{\mathrm{1}−{cos}\left({t}\right)}{\mathrm{2}}} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\:\frac{{dt}}{{a}\:+{a}\:{cos}\left(\mathrm{2}{t}\right)\:+\mathrm{1}\:−{cos}\left(\mathrm{2}{t}\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dt}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cos}\left(\mathrm{2}{t}\right)} \\ $$$$=_{\mathrm{2}{t}={x}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{4}\pi} \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cosx}}\:\frac{{dx}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cosx}}\:\:+\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cosx}} \\ $$$$\int_{\mathrm{2}\pi} ^{\mathrm{4}\pi} \:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cosx}}\:=_{{x}={t}+\mathrm{2}\pi} \:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cost}} \\ $$$$\Rightarrow\:{f}\left({a}\right)\:=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right){cosx}} \\ $$$${changement}\:\:{e}^{{ix}} \:={z}\:\:{give} \\ $$$${f}\left({a}\right)\:=\:\mathrm{2}\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{\mathrm{1}}{{a}+\mathrm{1}\:+\left({a}−\mathrm{1}\right)\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$${f}\left({a}\right)\:=\mathrm{4}\:\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{{dz}}{{iz}\left(\mathrm{2}+\mathrm{2}{a}\:+\left({a}−\mathrm{1}\right)\left({z}+{z}^{−\mathrm{1}} \right)\right)} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\:\frac{−\mathrm{4}{i}\:{dz}}{\mathrm{2}\left(\mathrm{1}+{a}\right){z}\:+\left({a}−\mathrm{1}\right){z}^{\mathrm{2}} \:+{a}−\mathrm{1}} \\ $$$${let}\:{inyroduce}\:{tbe}\:{complex}\:{function} \\ $$$$\varphi\left({z}\right)\:=\:\:\:\frac{−\mathrm{4}{i}}{\left({a}−\mathrm{1}\right){z}^{\mathrm{2}} \:\:+\mathrm{2}\left({a}+\mathrm{1}\right){z}\:+{a}−\mathrm{1}}\:.{poles}\:{of}\:\varphi? \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
Δ^′ = (a+1)^2 −(a−1)^2 = a^2 +2a+1−a^2 +2a−1=4a z_1 = ((−a−1 +2(√a))/(a−1)) = ((a −2(√a) +1)/(1−a)) = (((1−(√a))^2 )/((1−(√a))(1+(√a)))) = ((1−(√a))/(1+(√a))) and z_2 = ((−a−1−2(√a))/(a−1)) = ((a +2(√a) +1)/(1−a)) = (((1+(√a))^2 )/((1−(√a))(1+(√a)))) = ((1+(√a))/(1−(√a))) case1 a>1 ∣z_1 ∣ −1 = (((√a) −1)/( (√a)+1)) −1=(((√a) −1−(√a) −1)/( (√a) +1)) = ((−2)/( (√a) +1))<0 ⇒ ∣z_1 ∣<1 ∣z_2 ∣−1 = (1/(∣z_1 ∣)) −1 = ((1−∣z_1 ∣)/(∣z_1 ∣)) >0 ⇒ ∣z_2 ∣>1 (to eliminate from residus) ∫_(∣z∣=1) ϕ(2)dz =2iπ Res(ϕ,z_1 ) we have ϕ(z) = ((−4i)/((a−1)(z −z_1 )(z−z_2 ))) ⇒ Res(ϕ, z_1 ) = ((−4i)/((a−1)(z_1 −z_2 ))) = ((−4i)/((a−1)(z_1 −(1/z_1 )))) = ((−4i z_1 )/((a−1)(z_1 ^2 −1))) =((−4i ((1−(√a))/(1+(√a))))/((a−1)( (((1−(√a))^2 )/((1+(√a))^2 )) −1))) = ((−4i(1−(√a)))/((a−1)(1+(√a)){ (1−(√a))^2 −(1+(√a))^2 })) (1+(√a))^2 = ((4i)/(a−2(√a) +1 −a −2(√a) −1)) = ((4i)/(−4(√a))) =((−i)/( (√a))) ∫_(∣z∣=1) ϕ(z)dz =2iπ {((−i)/( (√a)))} = ((2π)/( (√a))) ⇒ f(a) = ((2π)/( (√a))) with a>1 case 2 0<a<1 ∣z_1 ∣ −1 =((1−(√a))/(1+(√a))) −1 = ((1−(√a)−1−(√a))/(1+(√a))) = ((−2(√a))/(1+a)) <0 ⇒ ∣z_1 ∣<1 and ∣z_2 ∣>1 so the value of f(a) dont change and f(a) = ((2π)/( (√a))) .
$$\Delta^{'} \:=\:\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left({a}−\mathrm{1}\right)^{\mathrm{2}} =\:{a}^{\mathrm{2}} \:+\mathrm{2}{a}+\mathrm{1}−{a}^{\mathrm{2}} \:+\mathrm{2}{a}−\mathrm{1}=\mathrm{4}{a} \\ $$$${z}_{\mathrm{1}} =\:\frac{−{a}−\mathrm{1}\:+\mathrm{2}\sqrt{{a}}}{{a}−\mathrm{1}}\:=\:\:\frac{{a}\:−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}}{\mathrm{1}−{a}}\:=\:\frac{\left(\mathrm{1}−\sqrt{{a}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\sqrt{{a}}\right)\left(\mathrm{1}+\sqrt{{a}}\right)} \\ $$$$=\:\frac{\mathrm{1}−\sqrt{{a}}}{\mathrm{1}+\sqrt{{a}}}\:\:\:{and}\:{z}_{\mathrm{2}} =\:\frac{−{a}−\mathrm{1}−\mathrm{2}\sqrt{{a}}}{{a}−\mathrm{1}}\:=\:\frac{{a}\:+\mathrm{2}\sqrt{{a}}\:+\mathrm{1}}{\mathrm{1}−{a}} \\ $$$$=\:\frac{\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} }{\left(\mathrm{1}−\sqrt{{a}}\right)\left(\mathrm{1}+\sqrt{{a}}\right)}\:=\:\frac{\mathrm{1}+\sqrt{{a}}}{\mathrm{1}−\sqrt{{a}}} \\ $$$${case}\mathrm{1}\:\:{a}>\mathrm{1}\:\:\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\:\frac{\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}+\mathrm{1}}\:−\mathrm{1}=\frac{\sqrt{{a}}\:−\mathrm{1}−\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}\:+\mathrm{1}} \\ $$$$=\:\frac{−\mathrm{2}}{\:\sqrt{{a}}\:+\mathrm{1}}<\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\:\frac{\mathrm{1}}{\mid{z}_{\mathrm{1}} \mid}\:−\mathrm{1}\:=\:\frac{\mathrm{1}−\mid{z}_{\mathrm{1}} \mid}{\mid{z}_{\mathrm{1}} \mid}\:>\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\left({to}\right. \\ $$$$\left.{eliminate}\:{from}\:{residus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left(\mathrm{2}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:\:{we}\:{have} \\ $$$$\varphi\left({z}\right)\:=\:\:\frac{−\mathrm{4}{i}}{\left({a}−\mathrm{1}\right)\left({z}\:−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,\:{z}_{\mathrm{1}} \right)\:=\:\:\:\frac{−\mathrm{4}{i}}{\left({a}−\mathrm{1}\right)\left({z}_{\mathrm{1}} \:−{z}_{\mathrm{2}} \right)} \\ $$$$=\:\frac{−\mathrm{4}{i}}{\left({a}−\mathrm{1}\right)\left({z}_{\mathrm{1}} \:−\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\right)}\:\:=\:\frac{−\mathrm{4}{i}\:{z}_{\mathrm{1}} }{\left({a}−\mathrm{1}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$$$=\frac{−\mathrm{4}{i}\:\frac{\mathrm{1}−\sqrt{{a}}}{\mathrm{1}+\sqrt{{a}}}}{\left({a}−\mathrm{1}\right)\left(\:\frac{\left(\mathrm{1}−\sqrt{{a}}\right)^{\mathrm{2}} }{\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} }\:−\mathrm{1}\right)} \\ $$$$=\:\frac{−\mathrm{4}{i}\left(\mathrm{1}−\sqrt{{a}}\right)}{\left({a}−\mathrm{1}\right)\left(\mathrm{1}+\sqrt{{a}}\right)\left\{\:\left(\mathrm{1}−\sqrt{{a}}\right)^{\mathrm{2}} \:−\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} \right\}}\:\left(\mathrm{1}+\sqrt{{a}}\right)^{\mathrm{2}} \\ $$$$=\:\frac{\mathrm{4}{i}}{{a}−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}\:−{a}\:−\mathrm{2}\sqrt{{a}}\:−\mathrm{1}}\:=\:\:\frac{\mathrm{4}{i}}{−\mathrm{4}\sqrt{{a}}}\:=\frac{−{i}}{\:\sqrt{{a}}} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\frac{−{i}}{\:\sqrt{{a}}}\right\}\:=\:\frac{\mathrm{2}\pi}{\:\sqrt{{a}}}\:\Rightarrow\: \\ $$$${f}\left({a}\right)\:=\:\:\frac{\mathrm{2}\pi}{\:\sqrt{{a}}}\:\:{with}\:\:{a}>\mathrm{1} \\ $$$${case}\:\mathrm{2}\:\:\:\:\:\mathrm{0}<{a}<\mathrm{1}\:\:\:\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\frac{\mathrm{1}−\sqrt{{a}}}{\mathrm{1}+\sqrt{{a}}}\:−\mathrm{1} \\ $$$$=\:\frac{\mathrm{1}−\sqrt{{a}}−\mathrm{1}−\sqrt{{a}}}{\mathrm{1}+\sqrt{{a}}}\:=\:\frac{−\mathrm{2}\sqrt{{a}}}{\mathrm{1}+{a}}\:<\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$${and}\:\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\:\:{so}\:{the}\:{value}\:{of}\:{f}\left({a}\right)\:{dont}\:{change}\:{and} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{2}\pi}{\:\sqrt{{a}}}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
we have f^′ (a) = ∫_0 ^(2π) (∂/∂a){ (1/(a cos^2 t +sin^2 t))}dt = − ∫_0 ^(2π) ((cos^2 t)/((acos^2 t +sin^2 t)^2 ))dt =−g(a) ⇒ g(a) =−f^′ (a) or f(a)= ((2π)/( (√a))) ⇒f^′ (a) =2π{ −((((√a))^′ )/a)} =−2π (1/(2a(√a))) = ((−π)/(a(√a))) ⇒ g(a) = −(π/(a(√a))) .
$${we}\:{have}\:{f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\partial}{\partial{a}}\left\{\:\:\:\:\:\:\frac{\mathrm{1}}{{a}\:{cos}^{\mathrm{2}} {t}\:+{sin}^{\mathrm{2}} {t}}\right\}{dt} \\ $$$$=\:−\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{cos}^{\mathrm{2}} {t}}{\left({acos}^{\mathrm{2}} {t}\:+{sin}^{\mathrm{2}} {t}\right)^{\mathrm{2}} }{dt}\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:\:{or}\:\:{f}\left({a}\right)=\:\frac{\mathrm{2}\pi}{\:\sqrt{{a}}}\:\Rightarrow{f}^{'} \left({a}\right)\:=\mathrm{2}\pi\left\{\:−\frac{\left(\sqrt{{a}}\right)^{'} }{{a}}\right\} \\ $$$$=−\mathrm{2}\pi\:\:\frac{\mathrm{1}}{\mathrm{2}{a}\sqrt{{a}}}\:=\:\frac{−\pi}{{a}\sqrt{{a}}}\:\Rightarrow\:\:{g}\left({a}\right)\:=\:−\frac{\pi}{{a}\sqrt{{a}}}\:. \\ $$

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