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Question Number 43909 by abdo.msup.com last updated on 17/Sep/18

find f(a,t)=∫_0 ^(2π) (dx/(a +t sinx)) 2)calculate ∫_0 ^(2π) ((sinx)/((a+tsinx)^2 ))dx 3)calculate ∫_0 ^(2π) (dx/((a+tsinx)^2 )) .

$${find}\:{f}\left({a},{t}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{a}\:+{t}\:{sinx}} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sinx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{3}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }\:. \\ $$

Answered by maxmathsup by imad last updated on 24/Sep/18

changement e^(ix) =z give f(a,t) = ∫_(∣z∣=1) (1/(a+t ((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((2i)/(iz{2ia +t(z−z^(−1) )})) dz = ∫_(∣z∣=1) (2/(2iaz +tz^2 −t))dz let ϕ(z) = (2/(tz^2 +2iaz −t))dz .poles of ϕ Δ^′ = −a^2 +t^2 =t^2 −a^2 case 1 Δ^′ >0 ⇒z_1 =((−ia +(√(t^2 −a^2 )))/t) and z_2 =((−ia−(√(t^2 −a^2 )))/t) ∣z_1 ∣ −1 =(1/(∣t∣)) (√(a^2 +t^2 −a^2 ))=1 ⇒∣z_1 ∣>1 ( out of circle) ∣z_2 ∣−1 = (1/(∣t∣))(√(a^2 +t^2 −a^2 ))=1 ⇒∣z_2 ∣>1 (out of circle) ⇒ ∫_(∣z∣) ϕ(z)dz =0 case 2 Δ^′ <0 ⇒t^2 −a^2 <0 ⇒ Δ^′ =(i(√(a^2 −t^2 )))^2 ⇒z_1 =((−ia +i(√(a^2 −t^2 )))/t) z_2 =((−ia −i(√(a^2 −t^2 )))/t) ∣z_1 ∣−1 = ((∣a−(√(a^2 −t^2 ))∣)/(∣t∣)) ∣z_2 ∣−1 =((∣a +(√(a^2 −t^2 ))∣)/(∣t∣)) if a>0 we get ∣z_1 ∣>1 and ∣z_2 ∣>1 ⇒ ∫_(∣z∣=1) ϕ(z)dz =0 if a<0 ∣z_1 ∣>1 and ∣z_2 ∣<1 ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_2 ) but ϕ(z)= (2/(t(z−z_1 )(z−z_2 ))) ⇒ Res(ϕ,z_2 )= (2/(t(z_2 −z_1 ))) =(2/(t(((−2i(√(a^2 −t^2 )))/t)))) =((−1)/(i(√(a^2 −t^2 )))) ⇒∫_(∣z∣=1) ϕ(z)dz =2iπ (((−1))/(i(√(a^2 −t^2 )))) =((−2π)/(√(a^2 −t^2 ))) =f(a,t)

$${changement}\:\:{e}^{{ix}} \:={z}\:{give}\:{f}\left({a},{t}\right)\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{\mathrm{1}}{{a}+{t}\:\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{i}}{{iz}\left\{\mathrm{2}{ia}\:+{t}\left({z}−{z}^{−\mathrm{1}} \right)\right\}}\:{dz}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{\mathrm{2}}{\mathrm{2}{iaz}\:+{tz}^{\mathrm{2}} \:−{t}}{dz} \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{2}}{{tz}^{\mathrm{2}} \:+\mathrm{2}{iaz}\:−{t}}{dz}\:\:.{poles}\:{of}\:\varphi\:\: \\ $$$$\Delta^{'} \:=\:−{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \:={t}^{\mathrm{2}} −{a}^{\mathrm{2}} \\ $$$${case}\:\mathrm{1}\:\:\:\:\:\:\:\:\Delta^{'} >\mathrm{0}\:\Rightarrow{z}_{\mathrm{1}} =\frac{−{ia}\:+\sqrt{{t}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{t}}\:\:\:{and}\:{z}_{\mathrm{2}} =\frac{−{ia}−\sqrt{{t}^{\mathrm{2}} −{a}^{\mathrm{2}} }}{{t}} \\ $$$$\mid{z}_{\mathrm{1}} \mid\:−\mathrm{1}\:=\frac{\mathrm{1}}{\mid{t}\mid}\:\sqrt{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid>\mathrm{1}\:\left(\:{out}\:{of}\:{circle}\right) \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\:\frac{\mathrm{1}}{\mid{t}\mid}\sqrt{{a}^{\mathrm{2}} \:+{t}^{\mathrm{2}} −{a}^{\mathrm{2}} }=\mathrm{1}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\left({out}\:{of}\:{circle}\right)\:\Rightarrow \\ $$$$\int_{\mid{z}\mid} \:\:\varphi\left({z}\right){dz}\:=\mathrm{0} \\ $$$${case}\:\mathrm{2}\:\:\:\:\Delta^{'} <\mathrm{0}\:\Rightarrow{t}^{\mathrm{2}} −{a}^{\mathrm{2}} <\mathrm{0}\:\Rightarrow\:\Delta^{'} =\left({i}\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =\frac{−{ia}\:+{i}\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{{t}} \\ $$$${z}_{\mathrm{2}} =\frac{−{ia}\:−{i}\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{{t}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\:\:\frac{\mid{a}−\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }\mid}{\mid{t}\mid}\:\: \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:\:=\frac{\mid{a}\:+\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }\mid}{\mid{t}\mid} \\ $$$${if}\:{a}>\mathrm{0}\:\:{we}\:{get}\:\mid{z}_{\mathrm{1}} \mid>\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid>\mathrm{1}\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{0} \\ $$$${if}\:{a}<\mathrm{0}\:\:\mid{z}_{\mathrm{1}} \mid>\mathrm{1}\:{and}\:\mid{z}_{\mathrm{2}} \mid<\mathrm{1}\:\Rightarrow\:\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right) \\ $$$${but}\:\varphi\left({z}\right)=\:\frac{\mathrm{2}}{{t}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right)=\:\frac{\mathrm{2}}{{t}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\:=\frac{\mathrm{2}}{{t}\left(\frac{−\mathrm{2}{i}\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}{{t}}\right)} \\ $$$$=\frac{−\mathrm{1}}{{i}\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:\Rightarrow\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\left(−\mathrm{1}\right)}{{i}\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:=\frac{−\mathrm{2}\pi}{\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:={f}\left({a},{t}\right) \\ $$$$ \\ $$

Commented by maxmathsup by imad last updated on 24/Sep/18

2) we have f(a,t) = ∫_0 ^(2π) (dx/(a+tsinx)) ⇒(∂f/∂t)(a,t) =−∫_0 ^(2π) ((sinx)/((a+tsinx)^2 ))dx ⇒ ∫_0 ^(2π) ((sinx)/((a +t sinx)^2 ))dx =−(∂f/∂t)(a,t) case1 ∣t∣>∣a∣ ⇒ f(a,t)=0 ⇒ ∫_0 ^(2π) ((sinx)/((a +tsinx)^2 ))dx=0 case 2 ∣t∣<∣a∣ ⇒f(a,t) =((−2π)/(√(a^2 −t^2 ))) ⇒(∂f/∂t)(a,t) =−2π { (a^2 −t^2 )^(−(1/2)) }^((1)/t) =−2π (−(1/2))(−2t) (a^2 −t^2 )^(−(3/2)) =((−2πt)/((a^2 −t^2 )(√(a^2 −t^2 )))) ⇒ ∫_0 ^(2π) ((sinx)/((a+tsinx)^2 ))dx = ((2πt)/((a^2 −t^2 )(√(a^2 −t^2 )))) .

$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({a},{t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{{a}+{tsinx}}\:\Rightarrow\frac{\partial{f}}{\partial{t}}\left({a},{t}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sinx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }{dx}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{sinx}}{\left({a}\:+{t}\:{sinx}\right)^{\mathrm{2}} }{dx}\:=−\frac{\partial{f}}{\partial{t}}\left({a},{t}\right) \\ $$$${case}\mathrm{1}\:\:\:\mid{t}\mid>\mid{a}\mid\:\:\:\Rightarrow\:{f}\left({a},{t}\right)=\mathrm{0}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{sinx}}{\left({a}\:+{tsinx}\right)^{\mathrm{2}} }{dx}=\mathrm{0} \\ $$$${case}\:\mathrm{2}\:\:\mid{t}\mid<\mid{a}\mid\:\Rightarrow{f}\left({a},{t}\right)\:=\frac{−\mathrm{2}\pi}{\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:\Rightarrow\frac{\partial{f}}{\partial{t}}\left({a},{t}\right)\:=−\mathrm{2}\pi\:\left\{\:\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{\left(\mathrm{1}\right)/{t}} \\ $$$$=−\mathrm{2}\pi\:\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(−\mathrm{2}{t}\right)\:\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:=\frac{−\mathrm{2}\pi{t}}{\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{sinx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }{dx}\:=\:\frac{\mathrm{2}\pi{t}}{\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:. \\ $$

Commented by maxmathsup by imad last updated on 24/Sep/18

3) we have f(a,t) = ∫_0 ^(2π) (dx/(a+tsinx)) ⇒(∂f/∂a)(a,t) =−∫_0 ^(2π) (dx/((a+tsinx)^2 )) ⇒ ∫_0 ^(2π) (dx/((a+tsinx)^2 )) =−(∂f/∂a)(a,t) case 1 ∣t∣>∣a∣ ⇒f(a,t)=0 ⇒ ∫_0 ^(2π) (dx/((a+tsinx)^2 )) =0 case 2 ∣t∣<∣a∣ ⇒f(a,t) =((−2π)/(√(a^2 −t^2 ))) ⇒(∂f/∂a)(a,t)=−2π{(a^2 −t^2 )^(−(1/2)) }^((1)/a) =−2π (2a)(−(1/2))(a^2 −t^2 )^(−(3/2)) =((2π)/((a^2 −t^2 )(√(a^2 −t^2 )))) ⇒ ∫_0 ^(2π) (dx/((a +t sinx)^2 )) =((−2π)/((a^2 −t^2 )(√(a^2 −t^2 )))) .

$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({a},{t}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{{a}+{tsinx}}\:\Rightarrow\frac{\partial{f}}{\partial{a}}\left({a},{t}\right)\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }\:=−\frac{\partial{f}}{\partial{a}}\left({a},{t}\right) \\ $$$${case}\:\mathrm{1}\:\:\mid{t}\mid>\mid{a}\mid\:\Rightarrow{f}\left({a},{t}\right)=\mathrm{0}\:\Rightarrow\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{dx}}{\left({a}+{tsinx}\right)^{\mathrm{2}} }\:=\mathrm{0} \\ $$$${case}\:\mathrm{2}\:\mid{t}\mid<\mid{a}\mid\:\Rightarrow{f}\left({a},{t}\right)\:=\frac{−\mathrm{2}\pi}{\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:\Rightarrow\frac{\partial{f}}{\partial{a}}\left({a},{t}\right)=−\mathrm{2}\pi\left\{\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \right\}^{\left(\mathrm{1}\right)/{a}} \\ $$$$=−\mathrm{2}\pi\:\left(\mathrm{2}{a}\right)\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:\:=\frac{\mathrm{2}\pi}{\left({a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\frac{{dx}}{\left({a}\:+{t}\:{sinx}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}\pi}{\left({a}^{\mathrm{2}} −{t}^{\mathrm{2}} \right)\sqrt{{a}^{\mathrm{2}} −{t}^{\mathrm{2}} }}\:. \\ $$

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