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Question Number 27295 by iy last updated on 04/Jan/18
The value of the integral ∫_( 0) ^π (1/(a^2 −2a cos x+1)) dx (a< 1) is
$$\mathrm{The}\:\mathrm{value}\:\mathrm{of}\:\mathrm{the}\:\mathrm{integral} \\ $$$$\underset{\:\mathrm{0}} {\overset{\pi} {\int}}\:\:\frac{\mathrm{1}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\:\mathrm{cos}\:{x}+\mathrm{1}}\:{dx}\:\:\left({a}<\:\mathrm{1}\right)\:\mathrm{is} \\ $$
Commented by abdo imad last updated on 04/Jan/18
let do the changement tan((x/2))=t I=∫_0 ^π (dx/(a^2 −2a cosx +1))= ∫_0 ^∝ ((dt/(1+t^2 ))/(a^2 −2a((1−t^2 )/(1+t^2 )) +1)) = ∫_0 ^∝ (dt/(a^2 (1+t^2 ) −2a(1−t^2 )+1+t^2 )) = ∫_0 ^∝ (dt/((a^2 +2a +1)t^2 +a^2 −2a +1)) = ∫_0 ^∝ (dt/((a+1)^2 t^2 +(a−1)^2 )) =∫_0 ^∝ (dt/((a+1)^2 ( t^2 + (((1−a)/(a+1)))^2 ))) and we do the changeent t= ((1−a)/(a+1)) α ⇒I= (1/((a+1)^2 )) ∫_0 ^∝ ((((1−a)/(a+1))dα)/((((1−a)/(a+1)))^2 (1+α^2 ))) =(1/((a+1)^2 ))((1−a)/(a+1))(((a+1)^2 )/((1−a)^2 )) (π/2) =(π/(2(1−a^2 ))) .
$${let}\:{do}\:{the}\:{changement}\:\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t} \\ $$$${I}=\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{dx}}{{a}^{\mathrm{2}} −\mathrm{2}{a}\:{cosx}\:+\mathrm{1}}=\:\int_{\mathrm{0}} ^{\propto} \frac{\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }}{{a}^{\mathrm{2}} −\mathrm{2}{a}\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }\:+\mathrm{1}} \\ $$$$ \\ $$$$=\:\int_{\mathrm{0}} ^{\propto} \frac{{dt}}{{a}^{\mathrm{2}} \left(\mathrm{1}+{t}^{\mathrm{2}} \right)\:−\mathrm{2}{a}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)+\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\propto} \:\:\:\:\:\:\frac{{dt}}{\left({a}^{\mathrm{2}} +\mathrm{2}{a}\:+\mathrm{1}\right){t}^{\mathrm{2}} \:+{a}^{\mathrm{2}} −\mathrm{2}{a}\:+\mathrm{1}} \\ $$$$=\:\int_{\mathrm{0}} ^{\propto} \:\:\:\:\frac{{dt}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{2}} \:+\left({a}−\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\propto} \:\:\:\:\:\frac{{dt}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} \left(\:{t}^{\mathrm{2}} \:+\:\left(\frac{\mathrm{1}−{a}}{{a}+\mathrm{1}}\right)^{\mathrm{2}} \:\right)}\:{and}\:{we}\:{do}\:{the}\:{changeent} \\ $$$${t}=\:\frac{\mathrm{1}−{a}}{{a}+\mathrm{1}}\:\alpha\:\Rightarrow{I}=\:\frac{\mathrm{1}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\:\int_{\mathrm{0}} ^{\propto} \:\:\frac{\frac{\mathrm{1}−{a}}{{a}+\mathrm{1}}{d}\alpha}{\left(\frac{\mathrm{1}−{a}}{{a}+\mathrm{1}}\right)^{\mathrm{2}} \left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{1}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }\frac{\mathrm{1}−{a}}{{a}+\mathrm{1}}\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} }{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}} \\ $$$$=\frac{\pi}{\mathrm{2}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:. \\ $$

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