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Question Number 72908 by mathmax by abdo last updated on 04/Nov/19
find ∫_0 ^π (dθ/(x^2 −2x cosθ +1)) with x real.
$${find}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{cos}\theta\:+\mathrm{1}}\:\:{with}\:{x}\:{real}. \\ $$
Commented by mind is power last updated on 05/Nov/19
this is True for ∣x∣<1
$$\mathrm{this}\:\mathrm{is}\:\mathrm{True}\:\mathrm{for}\:\mid\mathrm{x}\mid<\mathrm{1} \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
Commented by Tanmay chaudhury last updated on 05/Nov/19
(1/(1−x^2 ))∫_0 ^π ((1−x^2 )/(x^2 −2xcosθ+1))dθ [when x<1 ] (1/(1−x^2 ))∫_0 ^π (1+2xcosθ+2x^2 cos2θ+2x^3 cos3θ+...)dθ (1/(1−x^2 ))[∣θ∣_0 ^π +others terms=0 =(π/(1−x^2 )) answer case−II when x>1 let k=(1/x) when x>1 so k<1 ∫_0 ^π (dθ/(x^2 −2xcosθ+1))dθ ∫_0 ^π (dθ/((1/k^2 )−(2/k)cosθ+1))dθ ∫_0 ^π ((k^2 dθ)/(1−2kcosθ+k^2 ))dθ [note k<1] k^2 ×(π/(1−k^2 )) =(π/((1/k^2 )−1))=so answer is =(π/(x^2 −1))
$$\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{1}−{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}}{d}\theta\:\:\:\left[{when}\:{x}<\mathrm{1}\:\:\right] \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \left(\mathrm{1}+\mathrm{2}{xcos}\theta+\mathrm{2}{x}^{\mathrm{2}} {cos}\mathrm{2}\theta+\mathrm{2}{x}^{\mathrm{3}} {cos}\mathrm{3}\theta+…\right){d}\theta \\ $$$$\frac{\mathrm{1}}{\mathrm{1}−{x}^{\mathrm{2}} }\left[\mid\theta\mid_{\mathrm{0}} ^{\pi} +{others}\:{terms}=\mathrm{0}\right. \\ $$$$=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:{answer}\:\: \\ $$$${case}−{II}\:\:{when}\:{x}>\mathrm{1} \\ $$$${let}\:{k}=\frac{\mathrm{1}}{{x}}\:\:\:{when}\:{x}>\mathrm{1}\:\:\:{so}\:\:{k}<\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta+\mathrm{1}}{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\frac{\mathrm{2}}{{k}}{cos}\theta+\mathrm{1}}{d}\theta \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{{k}^{\mathrm{2}} {d}\theta}{\mathrm{1}−\mathrm{2}{kcos}\theta+{k}^{\mathrm{2}} }{d}\theta\:\:\:\left[\boldsymbol{{note}}\:\boldsymbol{{k}}<\mathrm{1}\right] \\ $$$$\boldsymbol{{k}}^{\mathrm{2}} ×\frac{\pi}{\mathrm{1}−{k}^{\mathrm{2}} } \\ $$$$=\frac{\pi}{\frac{\mathrm{1}}{{k}^{\mathrm{2}} }−\mathrm{1}}=\boldsymbol{{so}}\:\boldsymbol{{answer}}\:\boldsymbol{{is}}\:\:\:=\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}} \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
let f(x)=∫_0 ^π (dθ/(x^2 −2xcosθ +1)) changement tan((θ/2))=tgive f(x)=∫_0 ^∞ ((2dt)/((1+t^2 )(x^2 −2x((1−t^2 )/(1+t^2 )) +1))) =∫_0 ^∞ ((2dt)/(x^2 (1+t^2 )−2x(1−t^2 )+1+t^2 )) =∫_0 ^∞ ((2dt)/(x^2 +x^2 t^2 −2x+2xt^2 +1+t^2 )) =∫_0 ^∞ ((2dt)/((x^2 +2x+1)t^2 +x^2 −2x+1)) =∫_0 ^∞ ((2dt)/((x+1)^2 t^2 +(x−1)^2 )) =(1/((x+1)^2 ))∫_0 ^∞ ((2dt)/(t^2 +(((x−1)/(x+1)))^2 ))( if x≠−1 and x≠1) ⇒f(x)=_(t=∣((x−1)/(x+1))∣u) (1/((x+1)^2 ))∫_0 ^∞ (2/((((x−1)/(x+1)))^2 {1+u^2 }))∣((x−1)/(x+1))∣du =(1/((x−1)^2 ))×((∣x−1∣)/(∣x+1∣))∫_0 ^∞ ((2du)/(1+u^2 )) =(1/(∣x^2 −1∣))(2×(π/2))=(π/(∣x^2 −1∣)) ⇒ f(x)=(π/(∣x^2 −1∣))
$${let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{xcos}\theta\:+\mathrm{1}}\:\:{changement}\:{tan}\left(\frac{\theta}{\mathrm{2}}\right)={tgive} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} −\mathrm{2}{x}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{1}\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{{x}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)−\mathrm{2}{x}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{{x}^{\mathrm{2}} \:+{x}^{\mathrm{2}} {t}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{2}{xt}^{\mathrm{2}} \:+\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}\right){t}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{2}{x}+\mathrm{1}} \\ $$$$=\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{2}} +\left({x}−\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} \:+\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }\left(\:\:{if}\:{x}\neq−\mathrm{1}\:{and}\:\right. \\ $$$$\left.{x}\neq\mathrm{1}\right)\:\Rightarrow{f}\left({x}\right)=_{{t}=\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid{u}} \:\:\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}}{\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} \left\{\mathrm{1}+{u}^{\mathrm{2}} \right\}}\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\mid{du} \\ $$$$=\frac{\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }×\frac{\mid{x}−\mathrm{1}\mid}{\mid{x}+\mathrm{1}\mid}\int_{\mathrm{0}} ^{\infty} \:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}\left(\mathrm{2}×\frac{\pi}{\mathrm{2}}\right)=\frac{\pi}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\pi}{\mid{x}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
whom you target by this remarks...reply now
$${whom}\:{you}\:{target}\:{by}\:{this}\:{remarks}…{reply}\:{now} \\ $$
Commented by ajfour last updated on 05/Nov/19
Great method, Tanmay Sir..
$${Great}\:{method},\:{Tanmay}\:{Sir}.. \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
Thank you sir...
$${Thank}\:{you}\:{sir}… \\ $$
Commented by ajfour last updated on 05/Nov/19
sorry sir, i concluded this on a fleeting glance, forgive me..
$${sorry}\:{sir},\:{i}\:{concluded}\:{this}\:{on}\:{a} \\ $$$${fleeting}\:{glance},\:{forgive}\:{me}.. \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
Commented by Tanmay chaudhury last updated on 05/Nov/19
taken from SL loney
$${taken}\:{from}\:{SL}\:{loney} \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
taken from ML khanna
$${taken}\:{from}\:{ML}\:{khanna} \\ $$
Commented by Tanmay chaudhury last updated on 05/Nov/19
Commented by Tanmay chaudhury last updated on 05/Nov/19
sir pls check i have considerd both case x<1 and x>1 pls
$${sir}\:{pls}\:{check}\:{i}\:{have}\:{considerd}\:\:{both}\:{case} \\ $$$${x}<\mathrm{1}\:\:\:\:{and}\:\:{x}>\mathrm{1}\:\:{pls} \\ $$
Commented by mind is power last updated on 05/Nov/19
nice solution
$$\mathrm{nice}\:\mathrm{solution}\:\: \\ $$
Commented by mind is power last updated on 05/Nov/19
nice sir
$$\mathrm{nice}\:\mathrm{sir}\: \\ $$
Answered by mind is power last updated on 05/Nov/19
x^2 −2xcos(θ)+1 =(x−e^(iθ) )(x−e^(−iθ) ) ∫_0 ^π (dθ/((x−e^(iθ) )(x−e^(−iθ) )))=∫_0 (dθ/((x−e^(iθ) )(xe^(iθ) −1))) f(x)=∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))=f(−x) juste find f in x∈R_+ z=e^(iθ) ⇒e^(iθ) =−idz ⇒∫_1 ^(−1) ((−idz)/((x−z)(xz−1)))=i∫_1 ^(−1) (dz/((z−x)(xz−1))) =i∫_1 ^(−1) (1/((x^2 −1)(z−x)))dz+((xdz)/((1−x^2 )(xz−1))) =(i/(x^2 −1)){∫_1 ^(−1) (dz/(z−x))+∫_(−1) ^1 ((xdz)/(xz−1))} =(i/(x^2 −1))(ln(−1−x)−ln(1−x)+ln(x−1)−ln(−x−1) =(i/(x^2 −1))(ln(x−1)−ln(1−x)) ln(z_1 .z_2 )=ln(z_1 )+ln(z_2 )+2inπ if x>1⇒ln(1−x)=ln(−1(x−1)=iπ+ln(x−1) if x<1 ln(x−1)=iπ+ln(1−x) ⇒ ln(x−1)−ln(1−x)=−iπ x>1 ∫_0 ^π (dθ/(x^2 −2xcos(θ)+1))= { (((π/(x^2 −1)) x>1)),((((−π)/(x^2 −1)) 0≤ x<1)) :}
$$\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\theta\right)+\mathrm{1} \\ $$$$ \\ $$$$=\left(\mathrm{x}−\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{x}−\mathrm{e}^{−\mathrm{i}\theta} \right) \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{d}\theta}{\left(\mathrm{x}−\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{x}−\mathrm{e}^{−\mathrm{i}\theta} \right)}=\int_{\mathrm{0}} \frac{\mathrm{d}\theta}{\left(\mathrm{x}−\mathrm{e}^{\mathrm{i}\theta} \right)\left(\mathrm{xe}^{\mathrm{i}\theta} −\mathrm{1}\right)} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{d}\theta}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\theta\right)+\mathrm{1}}=\mathrm{f}\left(−\mathrm{x}\right) \\ $$$$\mathrm{juste}\:\mathrm{find}\:\mathrm{f}\:\mathrm{in}\:\mathrm{x}\in\mathbb{R}_{+} \\ $$$$\mathrm{z}=\mathrm{e}^{\mathrm{i}\theta} \Rightarrow\mathrm{e}^{\mathrm{i}\theta} =−\mathrm{idz} \\ $$$$\Rightarrow\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{−\mathrm{idz}}{\left(\mathrm{x}−\mathrm{z}\right)\left(\mathrm{xz}−\mathrm{1}\right)}=\mathrm{i}\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\mathrm{dz}}{\left(\mathrm{z}−\mathrm{x}\right)\left(\mathrm{xz}−\mathrm{1}\right)} \\ $$$$=\mathrm{i}\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{2}} −\mathrm{1}\right)\left(\mathrm{z}−\mathrm{x}\right)}\mathrm{dz}+\frac{\mathrm{xdz}}{\left(\mathrm{1}−\mathrm{x}^{\mathrm{2}} \right)\left(\mathrm{xz}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{i}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\left\{\int_{\mathrm{1}} ^{−\mathrm{1}} \frac{\mathrm{dz}}{\mathrm{z}−\mathrm{x}}+\int_{−\mathrm{1}} ^{\mathrm{1}} \frac{\mathrm{xdz}}{\mathrm{xz}−\mathrm{1}}\right\} \\ $$$$=\frac{\mathrm{i}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{ln}\left(−\mathrm{1}−\mathrm{x}\right)−\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)+\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{ln}\left(−\mathrm{x}−\mathrm{1}\right)\right. \\ $$$$=\frac{\mathrm{i}}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\left(\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)\right) \\ $$$$\mathrm{ln}\left(\mathrm{z}_{\mathrm{1}} .\mathrm{z}_{\mathrm{2}} \right)=\mathrm{ln}\left(\mathrm{z}_{\mathrm{1}} \right)+\mathrm{ln}\left(\mathrm{z}_{\mathrm{2}} \right)+\mathrm{2in}\pi \\ $$$$\mathrm{if}\:\mathrm{x}>\mathrm{1}\Rightarrow\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)=\mathrm{ln}\left(−\mathrm{1}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{i}\pi+\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)\right. \\ $$$$\mathrm{if}\:\mathrm{x}<\mathrm{1}\:\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)=\mathrm{i}\pi+\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right) \\ $$$$\Rightarrow \\ $$$$\mathrm{ln}\left(\mathrm{x}−\mathrm{1}\right)−\mathrm{ln}\left(\mathrm{1}−\mathrm{x}\right)=−\mathrm{i}\pi\:\:\:\mathrm{x}>\mathrm{1} \\ $$$$\int_{\mathrm{0}} ^{\pi} \frac{\mathrm{d}\theta}{\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\theta\right)+\mathrm{1}}=\begin{cases}{\frac{\pi}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\:\:\:\mathrm{x}>\mathrm{1}}\\{\frac{−\pi}{\mathrm{x}^{\mathrm{2}} −\mathrm{1}}\:\mathrm{0}\leqslant\:\mathrm{x}<\mathrm{1}}\end{cases} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by ajfour last updated on 05/Nov/19
I=∫_0 ^( π) (dθ/((x−1)^2 +4xsin^2 (θ/2))) dividing by cos^2 (θ/2) and let tan (θ/2)=t ⇒ (1/2)sec^2 (θ/2)dθ=dt ⇒ (sec^2 (θ/2))dθ=2dt I=∫^ ((2dt)/((x−1)^2 (1+t^2 )+4xt^2 )) =(1/((x+1)^2 ))∫((2dt)/(t^2 +(((x−1)/(x+1)))^2 ))=(2/(x^2 −1))tan^(−1) [((t(x+1))/(x−1))]+c =(2/(x^2 −1))((π/2)) = (π/(x^2 −1)) . I = (π/(x^2 −1)) ∙
$${I}=\int_{\mathrm{0}} ^{\:\pi} \frac{{d}\theta}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{4}{x}\mathrm{sin}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}} \\ $$$${dividing}\:{by}\:\mathrm{cos}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\:\:\:\:{and} \\ $$$$\:\:{let}\:\mathrm{tan}\:\frac{\theta}{\mathrm{2}}={t}\:\:\Rightarrow\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}{d}\theta={dt} \\ $$$$\Rightarrow\:\:\left(\mathrm{sec}\:^{\mathrm{2}} \frac{\theta}{\mathrm{2}}\right){d}\theta=\mathrm{2}{dt} \\ $$$${I}=\int^{\:} \frac{\mathrm{2}{dt}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)+\mathrm{4}{xt}^{\mathrm{2}} } \\ $$$$\:\:=\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }\int\frac{\mathrm{2}{dt}}{{t}^{\mathrm{2}} +\left(\frac{{x}−\mathrm{1}}{{x}+\mathrm{1}}\right)^{\mathrm{2}} }=\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\mathrm{tan}^{−\mathrm{1}} \left[\frac{{t}\left({x}+\mathrm{1}\right)}{{x}−\mathrm{1}}\right]+{c} \\ $$$$\:=\frac{\mathrm{2}}{{x}^{\mathrm{2}} −\mathrm{1}}\left(\frac{\pi}{\mathrm{2}}\right)\:=\:\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:. \\ $$$$\:{I}\:=\:\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:\centerdot \\ $$
Commented by mind is power last updated on 05/Nov/19
sir if ((x+1)/(x−1))<0 lim t→∞ tan^− ((((x+1}t)/(x−1)))=−(π/2)
$$\mathrm{sir}\:\mathrm{if}\:\frac{\mathrm{x}+\mathrm{1}}{\mathrm{x}−\mathrm{1}}<\mathrm{0} \\ $$$$\mathrm{lim}\:\mathrm{t}\rightarrow\infty\:\mathrm{tan}^{−} \left(\left(\frac{\left.\mathrm{x}+\mathrm{1}\right\}\mathrm{t}}{\mathrm{x}−\mathrm{1}}\right)=−\frac{\pi}{\mathrm{2}}\right. \\ $$
Commented by ajfour last updated on 05/Nov/19
true, thanks for the light.
$${true},\:{thanks}\:{for}\:{the}\:{light}. \\ $$

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