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Question Number 74513 by mathmax by abdo last updated on 25/Nov/19
find f(x)=∫_0 ^π (dθ/(x^2 −2xsin(2θ)+1)) (x real)
$${find}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xsin}\left(\mathrm{2}\theta\right)+\mathrm{1}}\:\:\left({x}\:{real}\right) \\ $$
Commented by mathmax by abdo last updated on 26/Nov/19
f(x)=_(2θ=t) ∫_0 ^(2π) (dt/(2(x^2 −2xsint +1))) ⇒2f(x)=∫_0 ^(2π) (dt/(x^2 −2xsint +1)) changement e^(it) =z give 2f(x)=∫_(∣z∣=1) (1/(x^2 −2x((z−z^(−1) )/(2i))+1))(dz/(iz)) =∫_(∣z∣=1) ((idz)/(iz(ix^2 −xz+xz^(−1) +i))) = =∫_(∣z∣=1) (dz/(ix^2 z−xz^2 +x +iz)) =∫_(∣z∣=1) (dz/(−xz^2 +i(1+x^2 )z +x)) =∫_(∣z∣=1) (dz/(−xz^2 +i(1+x^2 )z +x)) =−∫_(∣z∣=1) (dz/(xz^2 −i(1+x^2 )z −x)) let ϕ(z)=(1/(xz^2 −i(x^2 +1)z −x)) pole of ϕ? Δ=−(x^2 +1)^2 +4x^2 =4x^2 −(x^4 +2x^2 +1)=−x^4 +2x^2 −1 =−(x^4 −2x^2 +1) =−(x^2 −1)^2 = z_1 =((i(x^2 +1)+i∣x^2 −1∣)/(2x)) and z_2 =((i(x^2 +1)−i∣x^2 −1∣)/(2x)) case 1 ∣x∣<1 ⇒z_1 =((i(x^2 +1)+i(1−x^2 ))/(2x)) =(i/x) ∣z_1 ∣=(1/(∣x∣))>1 (out of circle) z_2 =((i(x^2 +1)−i(1−x^2 ))/(2x)) =((2ix^2 )/(2x)) =ix ⇒∣z_2 ∣=∣x∣<1 ϕ(z)=(1/(x(z−z_1 )(z−z_2 ))) and ∫_(∣z∣=1) ϕ(z)dz=2iπ Res(ϕ,z_2 ) =2iπ×(1/(x(z_2 −z_1 ))) =((2πi)/(x(ix−(i/x)))) =((2π)/((x^2 −1))) ⇒2f(x)=((−2π)/(x^2 −1)) ⇒ f(x)=(π/(1−x^2 )) with ∣x∣<1 case 2 ∣x∣>1 ⇒z_1 =((i(x^2 +1)+i(x^2 −1))/(2x)) =ix ⇒∣z_1 ∣=∣x∣>1(out of circle) z_2 =((i(x^2 +1)−i(x^2 −1))/(2x)) =(i/x) ⇒∣z_2 ∣=(1/(∣x∣))<1 ∫_(∣z∣=1) ϕ(z)dz =2iπRes(ϕ,z_2 ) =2iπ×(1/(x(z_2 −z_1 ))) =((2iπ)/(x((i/x)−ix))) =((2π)/(1−x^2 )) ⇒2f(x)=((−2π)/(1−x^2 )) ⇒f(x)=(π/(x^2 −1)) finally f(x)=(π/(1−x^2 )) if ∣x∣<1 and f(x)=(π/(x^2 −1)) if ∣x∣>1
$${f}\left({x}\right)=_{\mathrm{2}\theta={t}} \:\:\:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\:\:\frac{{dt}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{xsint}\:+\mathrm{1}\right)}\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{dt}}{{x}^{\mathrm{2}} −\mathrm{2}{xsint}\:+\mathrm{1}} \\ $$$${changement}\:\:{e}^{{it}} ={z}\:{give} \\ $$$$\mathrm{2}{f}\left({x}\right)=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\frac{{z}−{z}^{−\mathrm{1}} }{\mathrm{2}{i}}+\mathrm{1}}\frac{{dz}}{{iz}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\:\frac{{idz}}{{iz}\left({ix}^{\mathrm{2}} −{xz}+{xz}^{−\mathrm{1}} +{i}\right)}\:= \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{dz}}{{ix}^{\mathrm{2}} {z}−{xz}^{\mathrm{2}} +{x}\:+{iz}}\:=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{dz}}{−{xz}^{\mathrm{2}} +{i}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}} \\ $$$$=\int_{\mid{z}\mid=\mathrm{1}} \:\:\frac{{dz}}{−{xz}^{\mathrm{2}} +{i}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:+{x}}\:=−\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\frac{{dz}}{{xz}^{\mathrm{2}} −{i}\left(\mathrm{1}+{x}^{\mathrm{2}} \right){z}\:−{x}} \\ $$$${let}\:\varphi\left({z}\right)=\frac{\mathrm{1}}{{xz}^{\mathrm{2}} −{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right){z}\:−{x}}\:{pole}\:{of}\:\varphi? \\ $$$$\Delta=−\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} \:=\mathrm{4}{x}^{\mathrm{2}} −\left({x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)=−{x}^{\mathrm{4}} +\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=−\left({x}^{\mathrm{4}} −\mathrm{2}{x}^{\mathrm{2}} +\mathrm{1}\right)\:=−\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} = \\ $$$${z}_{\mathrm{1}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{i}\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{\mathrm{2}{x}}\:\:{and}\:\:{z}_{\mathrm{2}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{i}\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}{\mathrm{2}{x}} \\ $$$${case}\:\mathrm{1}\:\:\:\:\:\mid{x}\mid<\mathrm{1}\:\Rightarrow{z}_{\mathrm{1}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}}\:=\frac{{i}}{{x}} \\ $$$$\mid{z}_{\mathrm{1}} \mid=\frac{\mathrm{1}}{\mid{x}\mid}>\mathrm{1}\:\:\:\left({out}\:{of}\:{circle}\right) \\ $$$${z}_{\mathrm{2}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{i}\left(\mathrm{1}−{x}^{\mathrm{2}} \right)}{\mathrm{2}{x}}\:=\frac{\mathrm{2}{ix}^{\mathrm{2}} }{\mathrm{2}{x}}\:={ix}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid=\mid{x}\mid<\mathrm{1} \\ $$$$\varphi\left({z}\right)=\frac{\mathrm{1}}{{x}\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:{and}\:\int_{\mid{z}\mid=\mathrm{1}} \:\:\:\varphi\left({z}\right){dz}=\mathrm{2}{i}\pi\:{Res}\left(\varphi,{z}_{\mathrm{2}} \right) \\ $$$$=\mathrm{2}{i}\pi×\frac{\mathrm{1}}{{x}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)}\:=\frac{\mathrm{2}\pi{i}}{{x}\left({ix}−\frac{{i}}{{x}}\right)}\:=\frac{\mathrm{2}\pi}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)}\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\frac{−\mathrm{2}\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:{with}\:\mid{x}\mid<\mathrm{1} \\ $$$${case}\:\mathrm{2}\:\:\mid{x}\mid>\mathrm{1}\:\Rightarrow{z}_{\mathrm{1}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)+{i}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}}\:={ix}\:\Rightarrow\mid{z}_{\mathrm{1}} \mid=\mid{x}\mid>\mathrm{1}\left({out}\:{of}\right. \\ $$$$\left.{circle}\right)\:\:{z}_{\mathrm{2}} =\frac{{i}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−{i}\left({x}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}{x}}\:=\frac{{i}}{{x}}\:\Rightarrow\mid{z}_{\mathrm{2}} \mid=\frac{\mathrm{1}}{\mid{x}\mid}<\mathrm{1} \\ $$$$\int_{\mid{z}\mid=\mathrm{1}} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi{Res}\left(\varphi,{z}_{\mathrm{2}} \right)\:=\mathrm{2}{i}\pi×\frac{\mathrm{1}}{{x}\left({z}_{\mathrm{2}} −{z}_{\mathrm{1}} \right)} \\ $$$$=\frac{\mathrm{2}{i}\pi}{{x}\left(\frac{{i}}{{x}}−{ix}\right)}\:=\frac{\mathrm{2}\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{2}{f}\left({x}\right)=\frac{−\mathrm{2}\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:\Rightarrow{f}\left({x}\right)=\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$${finally}\:\:\:\:{f}\left({x}\right)=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:{if}\:\mid{x}\mid<\mathrm{1}\:{and}\:{f}\left({x}\right)=\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:{if}\:\mid{x}\mid>\mathrm{1} \\ $$
Answered by mind is power last updated on 25/Nov/19
f(x)=∫^π _0 (dθ/((x−sin(2θ))^2 +cos^2 (2θ))) x≠1 since 2−2sin(2θ) =0 for θ=(π/2) f(x)=f(−x) we suppose x≥0 =∫_0 ^π (dθ/((x−sin(2θ)−icos(θ))(x−sin(2θ)+icos(2θ)))) =∫_0 ^π (dθ/((x−ie^(−i2θ) )(x+ie^(i2θ) ))) z=e^(2iθ) ⇒dz=2ie^(2iθ) dθ =∫_C (dz/(2iz(x−(i/z))(x+iz)))=∫_C (dz/(2i(xz−i)(x+iz))) poles are (i/x),−ix if x=0 f(0)=π f(−1)=∫_0 ^π (dθ/(2+2sin(2θ)))=(dθ/(2(1+2sin(θ)cos(θ)))) =∫_0 ^(π/2) ((1+tg^2 (θ))/(2(1+tg(θ))^2 ))dθ+∫_(π/2) ^π =[−(1/(2(1+tg(θ)))]_0 ^(π/2) +[((−1)/(2(1+tg(θ))))]_(π/2) ^π =0 if ∣x∣>1⇒∣(i/x)∣<1∈C f(x)=π.Res((1/((xz−i)(x+iz))),z=(i/x)) =π.(x/(x^2 −1)) x>1 ∣x∣<1⇒∣−ix∣<1⇒f(x)=πRes((1/((xz−i)(x+iz))),z=ix} =(π/(1−x^2 )) ∣x∣<1
$${f}\left({x}\right)=\underset{\mathrm{0}} {\int}^{\pi} \frac{{d}\theta}{\left({x}−{sin}\left(\mathrm{2}\theta\right)\right)^{\mathrm{2}} +{cos}^{\mathrm{2}} \left(\mathrm{2}\theta\right)} \\ $$$${x}\neq\mathrm{1} \\ $$$${since}\:\mathrm{2}−\mathrm{2}{sin}\left(\mathrm{2}\theta\right)\:=\mathrm{0}\:{for}\:\theta=\frac{\pi}{\mathrm{2}} \\ $$$${f}\left({x}\right)={f}\left(−{x}\right)\:\:{we}\:{suppose}\:{x}\geqslant\mathrm{0} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{\left({x}−{sin}\left(\mathrm{2}\theta\right)−{icos}\left(\theta\right)\right)\left({x}−{sin}\left(\mathrm{2}\theta\right)+{icos}\left(\mathrm{2}\theta\right)\right)} \\ $$$$=\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{\left({x}−{ie}^{−{i}\mathrm{2}\theta} \right)\left({x}+{ie}^{{i}\mathrm{2}\theta} \right)} \\ $$$${z}={e}^{\mathrm{2}{i}\theta} \Rightarrow{dz}=\mathrm{2}{ie}^{\mathrm{2}{i}\theta} {d}\theta \\ $$$$=\int_{{C}} \frac{{dz}}{\mathrm{2}{iz}\left({x}−\frac{{i}}{{z}}\right)\left({x}+{iz}\right)}=\int_{{C}} \frac{{dz}}{\mathrm{2}{i}\left({xz}−{i}\right)\left({x}+{iz}\right)} \\ $$$${poles}\:{are}\:\frac{{i}}{{x}},−{ix} \\ $$$${if}\:\:{x}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)=\pi \\ $$$${f}\left(−\mathrm{1}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{\mathrm{2}+\mathrm{2}{sin}\left(\mathrm{2}\theta\right)}=\frac{{d}\theta}{\mathrm{2}\left(\mathrm{1}+\mathrm{2}{sin}\left(\theta\right){cos}\left(\theta\right)\right)} \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\mathrm{1}+{tg}^{\mathrm{2}} \left(\theta\right)}{\mathrm{2}\left(\mathrm{1}+{tg}\left(\theta\right)\right)^{\mathrm{2}} }{d}\theta+\int_{\frac{\pi}{\mathrm{2}}} ^{\pi} \\ $$$$=\left[−\frac{\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{tg}\left(\theta\right)\right.}\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\left[\frac{−\mathrm{1}}{\mathrm{2}\left(\mathrm{1}+{tg}\left(\theta\right)\right)}\right]_{\frac{\pi}{\mathrm{2}}} ^{\pi} =\mathrm{0} \\ $$$${if}\:\:\:\:\:\:\mid{x}\mid>\mathrm{1}\Rightarrow\mid\frac{{i}}{{x}}\mid<\mathrm{1}\in{C} \\ $$$${f}\left({x}\right)=\pi.{Res}\left(\frac{\mathrm{1}}{\left({xz}−{i}\right)\left({x}+{iz}\right)},{z}=\frac{{i}}{{x}}\right) \\ $$$$=\pi.\frac{{x}}{{x}^{\mathrm{2}} −\mathrm{1}}\:{x}>\mathrm{1} \\ $$$$\mid{x}\mid<\mathrm{1}\Rightarrow\mid−{ix}\mid<\mathrm{1}\Rightarrow{f}\left({x}\right)=\pi{Res}\left(\frac{\mathrm{1}}{\left({xz}−{i}\right)\left({x}+{iz}\right)},{z}={ix}\right\} \\ $$$$ \\ $$$$=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:\:\:\:\mid{x}\mid<\mathrm{1} \\ $$$$ \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 26/Nov/19
thankx sir.
$${thankx}\:{sir}. \\ $$
Commented by mind is power last updated on 26/Nov/19
y′re welcom
$${y}'{re}\:{welcom} \\ $$

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