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Question Number 46851 by maxmathsup by imad last updated on 01/Nov/18
let f(x)=∫_0 ^(2π) ((sint)/(x +sint))dt withx>1 1) calculate f(x) 2) calculate ∫_0 ^(2π) ((sint)/((x+sint)^2 ))dt 3)find the value of ∫_0 ^(2π) ((sint)/(2+sint))dt and ∫_0 ^(2π) ((sint)/((2+sint)^2 ))dt
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{{x}\:+{sint}}{dt}\:\:{withx}>\mathrm{1} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{{sint}}{\left({x}+{sint}\right)^{\mathrm{2}} }{dt} \\ $$$$\left.\mathrm{3}\right){find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\mathrm{2}+{sint}}{dt}\:{and}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left(\mathrm{2}+{sint}\right)^{\mathrm{2}} }{dt} \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
1) we have f(x)=∫_0 ^(2π) ((x+sint −x)/(x+sint)) dt =2π −x ∫_0 ^(2π) (dt/(x+sint)) changement e^(it) =z give ∫_0 ^(2π) (dt/(x+sint)) =∫_(∣z∣=1) (1/(x+((z−z^(−1) )/(2i)))) (dz/(iz)) = ∫_(∣z∣=1) ((2idz)/(iz(2ix +z−z^(−1) ))) = ∫_(∣z∣=1) ((2idz)/(−2xz+iz^2 −i)) =∫_(∣z∣=1) ((2dz)/(2ixz+z^2 −1)) =∫_(∣z∣=1) ((2dz)/(z^2 +2ixz −1)) let ϕ(z)=(2/(z^2 +2ixz −1)) poles of ϕ? Δ^′ =(ix)^2 +1 =1−x^2 =(i(√(x^2 −1)))^2 ⇒z_1 =−ix+i(√(x^2 −1)) z_2 =−ix−i(√(x^2 −1)) ∣z_1 ∣−1 =∣−x+(√(x^2 −1))∣−1 =(√(x^2 −1))−x−1=(√(x^2 −1))−(x+1) and x^2 −1−(x+1)^2 =x^2 −1−x^2 −2x−1 =−2x−2<0 ⇒∣z_1 ∣<1 ∣z_2 ∣−1 =∣x+(√(x^2 −1))∣−1 =x−1+(√(x^2 −1))>0 because x>1 ⇒z_2 is out of circle so ∫_(∣z∣=1) ϕ(z)dz =2iπ Res(ϕ,z_1 ) but ϕ(z) =(2/((z−z_1 )(z−z_2 ))) ⇒ Res(ϕ,z_1 ) =lim_(z→z_1 ) (z−z_1 )ϕ(z) =(2/(z_1 −z_2 )) =(2/(2i(√(x^2 −1)))) ⇒ ∫_(∣z∣=1) ϕ(z)dz =2iπ (1/(i(√(x^2 −1)))) =((2π)/( (√(x^2 −1)))) ⇒ ★f(x)=((2π)/( (√(x^2 −1)))) ★ with x>1
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{x}+{sint}\:−{x}}{{x}+{sint}}\:{dt}\:=\mathrm{2}\pi\:−{x}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{x}+{sint}} \\ $$$${changement}\:{e}^{{it}} ={z}\:{give}\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{x}+{sint}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{\mathrm{1}}{{x}+\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}}\:\frac{{dz}}{{iz}} \\ $$$$=\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{\mathrm{2}{idz}}{{iz}\left(\mathrm{2}{ix}\:+{z}−{z}^{−\mathrm{1}} \right)}\:=\:\int_{\mid{z}\mid=\mathrm{1}} \:\frac{\mathrm{2}{idz}}{−\mathrm{2}{xz}+{iz}^{\mathrm{2}} \:−{i}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\frac{\mathrm{2}{dz}}{\mathrm{2}{ixz}+{z}^{\mathrm{2}} −\mathrm{1}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\frac{\mathrm{2}{dz}}{{z}^{\mathrm{2}} \:+\mathrm{2}{ixz}\:−\mathrm{1}}\:\:{let}\:\varphi\left({z}\right)=\frac{\mathrm{2}}{{z}^{\mathrm{2}} \:+\mathrm{2}{ixz}\:−\mathrm{1}}\:{poles}\:{of}\:\varphi? \\ $$$$\Delta^{'} \:=\left({ix}\right)^{\mathrm{2}} \:+\mathrm{1}\:=\mathrm{1}−{x}^{\mathrm{2}} \:=\left({i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:\Rightarrow{z}_{\mathrm{1}} =−{ix}+{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${z}_{\mathrm{2}} =−{ix}−{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\mid{z}_{\mathrm{1}} \mid−\mathrm{1}\:=\mid−{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid−\mathrm{1}\:=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−{x}−\mathrm{1}=\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}−\left({x}+\mathrm{1}\right) \\ $$$${and}\:{x}^{\mathrm{2}} −\mathrm{1}−\left({x}+\mathrm{1}\right)^{\mathrm{2}} \:={x}^{\mathrm{2}} −\mathrm{1}−{x}^{\mathrm{2}} −\mathrm{2}{x}−\mathrm{1}\:=−\mathrm{2}{x}−\mathrm{2}<\mathrm{0}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid<\mathrm{1} \\ $$$$\mid{z}_{\mathrm{2}} \mid−\mathrm{1}\:=\mid{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\mid−\mathrm{1}\:={x}−\mathrm{1}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}>\mathrm{0}\:{because}\:{x}>\mathrm{1}\:\:\Rightarrow{z}_{\mathrm{2}} {is}\:{out}\:{of}\:{circle} \\ $$$${so}\:\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:{but}\:\varphi\left({z}\right)\:=\frac{\mathrm{2}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${Res}\left(\varphi,{z}_{\mathrm{1}} \right)\:={lim}_{{z}\rightarrow{z}_{\mathrm{1}} } \left({z}−{z}_{\mathrm{1}} \right)\varphi\left({z}\right)\:=\frac{\mathrm{2}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:=\frac{\mathrm{2}}{\mathrm{2}{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\frac{\mathrm{1}}{{i}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{\mathrm{2}\pi}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow\:\bigstar{f}\left({x}\right)=\frac{\mathrm{2}\pi}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\bigstar\:\:{with}\:{x}>\mathrm{1} \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
2) we have f^′ (x) =∫_0 ^(2π) (∂/∂x){ ((sint)/(x+sint))}dt =−∫_0 ^(2π) ((sint)/((x+sint)^2 ))dt ⇒ ∫_0 ^(2π) ((sint)/((x+sint)^2 ))dt =−f^′ (x) but f(x) =((2π)/( (√(x^2 −1)))) =2π(x^2 −1)^(−(1/2)) ⇒ f^′ (x) =2π(−(1/2))(2x)(x^2 −1)^(−(3/2)) =((−2πx)/((x^2 −1)^(3/2) )) =((−2πx)/((x^2 −1)(√(x^2 −1)))) ⇒ ∫_0 ^(2π) ((sint)/((x+sint)^2 )) dt =((2πx)/((x^2 −1)(√(x^2 −1)))) with x>1 .
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}^{'} \left({x}\right)\:=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\partial}{\partial{x}}\left\{\:\frac{{sint}}{{x}+{sint}}\right\}{dt}\:=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left({x}+{sint}\right)^{\mathrm{2}} }{dt}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left({x}+{sint}\right)^{\mathrm{2}} }{dt}\:=−{f}^{'} \left({x}\right)\:{but}\:{f}\left({x}\right)\:=\frac{\mathrm{2}\pi}{\:\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:=\mathrm{2}\pi\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \:\Rightarrow \\ $$$${f}^{'} \left({x}\right)\:=\mathrm{2}\pi\left(−\frac{\mathrm{1}}{\mathrm{2}}\right)\left(\mathrm{2}{x}\right)\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{−\frac{\mathrm{3}}{\mathrm{2}}} \:=\frac{−\mathrm{2}\pi{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} }\:=\frac{−\mathrm{2}\pi{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{{sint}}{\left({x}+{sint}\right)^{\mathrm{2}} }\:{dt}\:=\frac{\mathrm{2}\pi{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:{with}\:{x}>\mathrm{1}\:. \\ $$
Commented by maxmathsup by imad last updated on 02/Nov/18
3) we have ∫_0 ^(2π) ((sint)/(2+sint)) dt =f(2) =((2π)/( (√(2^2 −1)))) =((2π)/( (√3))) ∫_0 ^(2π) ((sint)/((2+sint)^2 ))dt =f^′ (2) =((4π)/((2^2 −1)(√(2^2 −1)))) =((4π)/(3(√3))) .
$$\left.\mathrm{3}\right)\:{we}\:{have}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\mathrm{2}+{sint}}\:{dt}\:={f}\left(\mathrm{2}\right)\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{\mathrm{2}\pi}{\:\sqrt{\mathrm{3}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{sint}}{\left(\mathrm{2}+{sint}\right)^{\mathrm{2}} }{dt}\:={f}^{'} \left(\mathrm{2}\right)\:=\frac{\mathrm{4}\pi}{\left(\mathrm{2}^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\mathrm{2}^{\mathrm{2}} −\mathrm{1}}}\:=\frac{\mathrm{4}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:. \\ $$

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