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Question Number 74514 by mathmax by abdo last updated on 25/Nov/19

calculate ∫_0 ^(2π) (((x−sinθ)dθ)/((x^2 −2x sinθ +1)^2 ))

$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\:\frac{\left({x}−{sin}\theta\right){d}\theta}{\left({x}^{\mathrm{2}} −\mathrm{2}{x}\:{sin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} } \\ $$

Commented by mathmax by abdo last updated on 26/Nov/19

we have proved that ∫_0 ^π (dt/(x^2 −2xsin(2t)+1)) =(π/(1−x^2 )) if ∣x∣<1 and =(π/(x^2 −1)) if ∣x∣>1 let f(x)=∫_0 ^π (dt/(x^2 −2x sin(2t)+1)) ⇒ f(x)=_(2t =θ) ∫_0 ^(2π) (dθ/(2(x^2 −2xsinθ +1))) ⇒2f^′ (x)=−∫_0 ^(2π) ((2x−2sinθ)/((x^2 (2xsinθ +1)^2 ))dθ ⇒ ∫_0 ^(2π) (((x−sinθ))/((x^2 −2xsinθ +1)^2 ))dθ =−f^′ (x) ∣x∣<1 ⇒f^′ (x)=−π((−2x)/((1−x^2 )^2 )) =((2πx)/((1−x^2 )^2 )) ∣x∣>1 ⇒f^′ (x)=−((π (2x))/((x^2 −1)^2 )) =((−2πx)/((x^2 −1)^2 )) so the value of this integral is known.

$${we}\:{have}\:{proved}\:{that}\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{dt}}{{x}^{\mathrm{2}} −\mathrm{2}{xsin}\left(\mathrm{2}{t}\right)+\mathrm{1}}\:=\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} }\:{if}\:\mid{x}\mid<\mathrm{1}\:{and} \\ $$$$=\frac{\pi}{{x}^{\mathrm{2}} −\mathrm{1}}\:{if}\:\mid{x}\mid>\mathrm{1}\:\:{let}\:{f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \:\frac{{dt}}{{x}^{\mathrm{2}} −\mathrm{2}{x}\:{sin}\left(\mathrm{2}{t}\right)+\mathrm{1}}\:\Rightarrow \\ $$$${f}\left({x}\right)=_{\mathrm{2}{t}\:=\theta} \:\:\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\:\frac{{d}\theta}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)}\:\Rightarrow\mathrm{2}{f}^{'} \left({x}\right)=−\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\mathrm{2}{x}−\mathrm{2}{sin}\theta}{\left({x}^{\mathrm{2}} \left(\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} \right.}{d}\theta\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\frac{\left({x}−{sin}\theta\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\theta\:+\mathrm{1}\right)^{\mathrm{2}} }{d}\theta\:=−{f}^{'} \left({x}\right) \\ $$$$\mid{x}\mid<\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=−\pi\frac{−\mathrm{2}{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} }\:=\frac{\mathrm{2}\pi{x}}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\mid{x}\mid>\mathrm{1}\:\Rightarrow{f}^{'} \left({x}\right)=−\frac{\pi\:\left(\mathrm{2}{x}\right)}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:=\frac{−\mathrm{2}\pi{x}}{\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }\:{so}\:{the}\:{value}\:{of}\:{this}\:{integral} \\ $$$${is}\:{known}. \\ $$

Answered by mind is power last updated on 25/Nov/19

use provious Quation f(x)=∫_0 ^π (dθ/(x^2 −2sin(2θ)x+1)) u=2θ f(x)=∫_0 ^(2π) (du/(2(x^2 −2sin(u)x+1))) ⇒2f(x)=∫_0 ^(2π) (du/(x^2 −2sin(u)x+1))= { ((((πx)/(x^2 −1)), x>1)),(((π/(1−x^2 )), 0<x<1)) :} 2f′(x)=∫_0 ^(2π) ((−2x+2sin(u))/((x^2 −2xsin(u)+1)^2 )) ⇒−f′(x)=∫_0 ^(2π) ((x−sin(u))/((x^2 −2xsin(u)+1)^2 ))du f′(x)=π.(((−x^2 −1)/((x^2 +1)^2 ))) 0≤x<1 f′(x)=((2xπ)/((1−x^2 )^2 )),x>1

$${use}\:{provious}\:{Quation} \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\pi} \frac{{d}\theta}{{x}^{\mathrm{2}} −\mathrm{2}{sin}\left(\mathrm{2}\theta\right){x}+\mathrm{1}} \\ $$$${u}=\mathrm{2}\theta \\ $$$${f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{\mathrm{2}\left({x}^{\mathrm{2}} −\mathrm{2}{sin}\left({u}\right){x}+\mathrm{1}\right)} \\ $$$$\Rightarrow\mathrm{2}{f}\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{du}}{{x}^{\mathrm{2}} −\mathrm{2}{sin}\left({u}\right){x}+\mathrm{1}}=\begin{cases}{\frac{\pi{x}}{{x}^{\mathrm{2}} −\mathrm{1}},\:\:{x}>\mathrm{1}}\\{\frac{\pi}{\mathrm{1}−{x}^{\mathrm{2}} },\:\:\:\mathrm{0}<{x}<\mathrm{1}}\end{cases} \\ $$$$\mathrm{2}{f}'\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{−\mathrm{2}{x}+\mathrm{2}{sin}\left({u}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\left({u}\right)+\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow−{f}'\left({x}\right)=\int_{\mathrm{0}} ^{\mathrm{2}\pi} \frac{{x}−{sin}\left({u}\right)}{\left({x}^{\mathrm{2}} −\mathrm{2}{xsin}\left({u}\right)+\mathrm{1}\right)^{\mathrm{2}} }{du} \\ $$$${f}'\left({x}\right)=\pi.\left(\frac{−{x}^{\mathrm{2}} −\mathrm{1}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{2}} }\right)\:\:\:\:\:\:\mathrm{0}\leqslant{x}<\mathrm{1} \\ $$$${f}'\left({x}\right)=\frac{\mathrm{2}{x}\pi}{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)^{\mathrm{2}} },{x}>\mathrm{1} \\ $$

Commented by mathmax by abdo last updated on 26/Nov/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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