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Question Number 35226 by abdo mathsup 649 cc last updated on 16/May/18
1) calculate f(a) = ∫_0 ^π (dx/(a sin^2 x +cos^2 x)) with a>0 2) find the value of g(a) = ∫_0 ^π ((sin^2 x)/((a sin^2 x +cos^2 x)^2 ))dx
$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\frac{{dx}}{{a}\:{sin}^{\mathrm{2}} {x}\:\:+{cos}^{\mathrm{2}} {x}} \\ $$$${with}\:{a}>\mathrm{0} \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{value}\:{of}\:{g}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{sin}^{\mathrm{2}} {x}}{\left({a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{dx} \\ $$
Commented by prof Abdo imad last updated on 19/May/18
1)we have f(a) = ∫_0 ^π (dx/(a sin^2 x +cos^2 x)) = ∫_0 ^π (dx/(a ((1−cos(2x))/2)+((1+cos(2x))/2))) = 2 ∫_0 ^π (dx/(a −a cos(2x) +1+cos(2x))) =2 ∫_0 ^π (dx/(a+1 +(1−a) cos(2x))) =_(2x =t) 2 ∫_0 ^(2π) (1/(1+a +(1−a)cost)) (dt/2) = ∫_0 ^(2π) (dt/(1+a +(1−a)cost)) changement e^(it) = z give I = ∫_(∣z∣=1) (1/(1+a +(1−a) ((z+z^(−1) )/2))) (dz/(iz)) I = ∫_(∣z∣=1) ((2dz)/(iz( 2+2a +(1−a)(z+z^(−1) ))) = ∫_(∣z∣=1) ((−2idz)/(2(1+a)z +(1−a)z^2 +1−a)) let consider ϕ(z) = ((−2i)/((1−a)^ z^2 +2(1+a)z +1−a)) poles of ϕ ?
$$\left.\mathrm{1}\right){we}\:{have}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{{a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}} \\ $$$$=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{{a}\:\frac{\mathrm{1}−{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}+\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\:\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{{a}\:−{a}\:{cos}\left(\mathrm{2}{x}\right)\:+\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=\mathrm{2}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\:\:\frac{{dx}}{{a}+\mathrm{1}\:+\left(\mathrm{1}−{a}\right)\:{cos}\left(\mathrm{2}{x}\right)} \\ $$$$=_{\mathrm{2}{x}\:={t}} \:\:\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cost}}\:\frac{{dt}}{\mathrm{2}} \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\:\:\:\frac{{dt}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right){cost}}\:\:{changement} \\ $$$${e}^{{it}} \:=\:{z}\:\:{give}\: \\ $$$${I}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{a}\:+\left(\mathrm{1}−{a}\right)\:\frac{{z}+{z}^{−\mathrm{1}} }{\mathrm{2}}}\:\frac{{dz}}{{iz}} \\ $$$${I}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{2}{dz}}{{iz}\left(\:\mathrm{2}+\mathrm{2}{a}\:+\left(\mathrm{1}−{a}\right)\left({z}+{z}^{−\mathrm{1}} \right)\right.} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{−\mathrm{2}{idz}}{\mathrm{2}\left(\mathrm{1}+{a}\right){z}\:\:+\left(\mathrm{1}−{a}\right){z}^{\mathrm{2}} \:\:+\mathrm{1}−{a}} \\ $$$${let}\:{consider}\:\varphi\left({z}\right)\:=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)^{} {z}^{\mathrm{2}} \:\:+\mathrm{2}\left(\mathrm{1}+{a}\right){z}\:+\mathrm{1}−{a}} \\ $$$${poles}\:{of}\:\varphi\:? \\ $$
Commented by prof Abdo imad last updated on 19/May/18
Δ^′ =(1+a)^2 −(1−a)^2 =1+2a +a^2 −1 +2a −a^2 =4a ⇒z_1 =((−1−a +2(√a))/(1−a)) =((a−2(√a) +1)/(a−1)) = ((((√a) −1)^2 )/(((√a) −1)((√a) +1))) = (((√a) −1)/( (√a) +1)) z_2 = ((−1−a −2(√a))/(1−a)) = ((((√a) +1)^2 )/(a−1)) =((((√a) +1)^2 )/(((√a) −1)((√a) +1))) = (((√a) +1)/( (√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 elominate from rrsidus) ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ, z_1 ) Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) but ϕ(z) = ((−2i)/((1−a)(z−z_1 )(z−z_2 ))) Res(ϕ,z_1 )= ((−2i)/((1−a)(z_1 −z_2 ))) = ((−2i)/((1−a)(z_1 −(1/z_1 )))) = ((−2i z_1 )/((1−a)(z_1 ^2 −1))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ ((−2iz_1 )/((1−a)(z_1 ^2 −1))) = ((4πz_1 )/((1−a)(z_1 ^2 −1)))
$$\Delta^{'} \:\:\:=\left(\mathrm{1}+{a}\right)^{\mathrm{2}} \:−\left(\mathrm{1}−{a}\right)^{\mathrm{2}} =\mathrm{1}+\mathrm{2}{a}\:+{a}^{\mathrm{2}} \:−\mathrm{1}\:+\mathrm{2}{a}\:−{a}^{\mathrm{2}} \\ $$$$=\mathrm{4}{a}\:\:\Rightarrow{z}_{\mathrm{1}} =\frac{−\mathrm{1}−{a}\:+\mathrm{2}\sqrt{{a}}}{\mathrm{1}−{a}}\:=\frac{{a}−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}}{{a}−\mathrm{1}} \\ $$$$=\:\frac{\left(\sqrt{{a}}\:−\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:−\mathrm{1}\right)\left(\sqrt{{a}}\:+\mathrm{1}\right)}\:=\:\frac{\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}\:\:+\mathrm{1}} \\ $$$${z}_{\mathrm{2}} \:=\:\frac{−\mathrm{1}−{a}\:−\mathrm{2}\sqrt{{a}}}{\mathrm{1}−{a}}\:=\:\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{{a}−\mathrm{1}}\:=\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:−\mathrm{1}\right)\left(\sqrt{{a}}\:+\mathrm{1}\right)} \\ $$$$=\:\frac{\sqrt{{a}}\:+\mathrm{1}}{\:\sqrt{{a}}\:−\mathrm{1}} \\ $$$$\:\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\:\frac{\sqrt{{a}^{} }−\mathrm{1}}{\:\sqrt{{a}}+\mathrm{1}}−\mathrm{1}=\:\frac{\sqrt{{a}}\:−\mathrm{1}\:−\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}\:+\mathrm{1}} \\ $$$$=\:\frac{−\mathrm{2}}{\:\sqrt{{a}}\:+\mathrm{1}}\:<\mathrm{0}\Rightarrow\:\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid\:−\mathrm{1}\:=\:\:\frac{\mathrm{1}}{\mid{z}_{\mathrm{1}} \mid}\:−\mathrm{1}\:=\:\frac{\mathrm{1}−\mid{z}_{\mathrm{1}} \mid}{\mid{z}_{\mathrm{1}} \mid}\:>\mathrm{0}\:\Rightarrow\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\left({to}\right. \\ $$$$\left.{elominate}\:{from}\:{rrsidus}\right) \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,\:{z}_{\mathrm{1}} \right) \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right)\varphi\left({z}\right) \\ $$$${but}\:\varphi\left({z}\right)\:=\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)} \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)=\:\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} −{z}_{\mathrm{2}} \right)}\:=\:\frac{−\mathrm{2}{i}}{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} \:−\frac{\mathrm{1}}{{z}_{\mathrm{1}} }\right)} \\ $$$$=\:\:\frac{−\mathrm{2}{i}\:{z}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\:\frac{−\mathrm{2}{iz}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{4}\pi{z}_{\mathrm{1}} }{\left(\mathrm{1}−{a}\right)\left({z}_{\mathrm{1}} ^{\mathrm{2}} \:−\mathrm{1}\right)} \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
∫_(−∞) ^(+∞) ϕ(z)dz = ((4π)/(1−a)) ((((√a) −1)/( (√a) +1))/(((((√a)−1)^2 )/(((√a) +1)^2 )) −1)) = ((4π)/(1−a)) (((√a) −1)/( (√a) +1)) ((((√a) +1)^2 )/(((√a)−1)^2 −((√a) +1)^2 )) = ((4π(a−1))/((1−a)( a −2(√a) +1 −a−2(√a) −1))) = ((4π)/(4(√a))) = (π/( (√a))) so f(a) = (π/( (√a))) .
$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\:\frac{\mathrm{4}\pi}{\mathrm{1}−{a}}\:\:\frac{\frac{\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}\:+\mathrm{1}}}{\frac{\left(\sqrt{{a}}−\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }\:−\mathrm{1}} \\ $$$$=\:\frac{\mathrm{4}\pi}{\mathrm{1}−{a}}\:\:\frac{\sqrt{{a}}\:−\mathrm{1}}{\:\sqrt{{a}}\:+\mathrm{1}}\:\frac{\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} }{\left(\sqrt{{a}}−\mathrm{1}\right)^{\mathrm{2}} \:−\left(\sqrt{{a}}\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{4}\pi\left({a}−\mathrm{1}\right)}{\left(\mathrm{1}−{a}\right)\left(\:{a}\:−\mathrm{2}\sqrt{{a}}\:+\mathrm{1}\:−{a}−\mathrm{2}\sqrt{{a}}\:−\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{4}\pi}{\mathrm{4}\sqrt{{a}}}\:=\:\frac{\pi}{\:\sqrt{{a}}}\:\:{so} \\ $$$${f}\left({a}\right)\:=\:\frac{\pi}{\:\sqrt{{a}}}\:. \\ $$
Commented by abdo mathsup 649 cc last updated on 20/May/18
2) we have f^′ (a) = ∫_0 ^π (∂/∂a){ (1/(asin^2 x +cos^2 x))}dx = −∫_0 ^π ((sin^2 x)/((a sin^2 x +cos^2 x)^2 ))dx =−g(a) ⇒ g(a) =−f^′ (a) but f(a)= (π/( (√a))) ⇒ f^′ (a)= −π ((((√a))^′ )/a) =−π (1/(2a(√a))) ⇒ g(a) = ((−π)/(2a(√a))) .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{\partial}{\partial{a}}\left\{\:\:\:\frac{\mathrm{1}}{{asin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}}\right\}{dx} \\ $$$$=\:−\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{sin}^{\mathrm{2}} {x}}{\left({a}\:{sin}^{\mathrm{2}} {x}\:+{cos}^{\mathrm{2}} {x}\right)^{\mathrm{2}} }{dx}\:=−{g}\left({a}\right)\:\Rightarrow \\ $$$${g}\left({a}\right)\:=−{f}^{'} \left({a}\right)\:\:{but}\:\:\:{f}\left({a}\right)=\:\frac{\pi}{\:\sqrt{{a}}}\:\Rightarrow\:{f}^{'} \left({a}\right)=\:−\pi\:\frac{\left(\sqrt{{a}}\right)^{'} }{{a}} \\ $$$$=−\pi\:\:\frac{\mathrm{1}}{\mathrm{2}{a}\sqrt{{a}}}\:\Rightarrow\:\:{g}\left({a}\right)\:=\:\frac{−\pi}{\mathrm{2}{a}\sqrt{{a}}}\:\:. \\ $$

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