Question Number 226771 by Spillover last updated on 14/Dec/25
Answered by Frix last updated on 14/Dec/25
![a≠±1 We can say a>0 because ∫_0 ^π (dθ/(1−2acos θ +a^2 ))=∫_0 ^π (dθ/(1+2acos θ +a^2 )) If 0<a<1 I=∫_0 ^π (dθ/(1−2acos θ +a^2 )) Let b=(1/a) ⇒ 1<b<∞ I=b^2 ∫_0 ^π (dθ/(1−2bcos θ +b^2 )) ⇒ I>0 ⇒ I=(π/(∣a^2 −1∣)) I=∫_0 ^π (dθ/(1−2acos θ +a^2 )) =^([t=tan (θ/2)]) =2∫_0 ^∞ (dt/((a+1)^2 t^2 +(a−1)^2 ))= =(2/(∣a^2 −1∣))[tan^(−1) (∣((a+1)/(a−1))∣t)]_0 ^∞ =(π/(∣a^2 −1∣)) [We know ∫(du/(αu^2 +β))=((tan^(−1) ((u(√α))/( (√β))))/( (√α)(√β))) and (√((a±1)^2 ))=∣a±1∣]](Q226781.png)
$${a}\neq\pm\mathrm{1} \\ $$$$\mathrm{We}\:\mathrm{can}\:\mathrm{say}\:{a}>\mathrm{0}\:\mathrm{because} \\ $$$$\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} }=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}+\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} } \\ $$$$\mathrm{If}\:\mathrm{0}<{a}<\mathrm{1} \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} } \\ $$$$\mathrm{Let}\:{b}=\frac{\mathrm{1}}{{a}}\:\Rightarrow\:\mathrm{1}<{b}<\infty \\ $$$${I}={b}^{\mathrm{2}} \underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{b}\mathrm{cos}\:\theta\:+{b}^{\mathrm{2}} } \\ $$$$\Rightarrow\:{I}>\mathrm{0} \\ $$$$\Rightarrow\:{I}=\frac{\pi}{\mid{a}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$$$ \\ $$$${I}=\underset{\mathrm{0}} {\overset{\pi} {\int}}\frac{{d}\theta}{\mathrm{1}−\mathrm{2}{a}\mathrm{cos}\:\theta\:+{a}^{\mathrm{2}} }\:\overset{\left[{t}=\mathrm{tan}\:\frac{\theta}{\mathrm{2}}\right]} {=} \\ $$$$=\mathrm{2}\underset{\mathrm{0}} {\overset{\infty} {\int}}\frac{{dt}}{\left({a}+\mathrm{1}\right)^{\mathrm{2}} {t}^{\mathrm{2}} +\left({a}−\mathrm{1}\right)^{\mathrm{2}} }= \\ $$$$=\frac{\mathrm{2}}{\mid{a}^{\mathrm{2}} −\mathrm{1}\mid}\left[\mathrm{tan}^{−\mathrm{1}} \:\left(\mid\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}\mid{t}\right)\right]_{\mathrm{0}} ^{\infty} =\frac{\pi}{\mid{a}^{\mathrm{2}} −\mathrm{1}\mid} \\ $$$$ \\ $$$$ \\ $$$$\left[\mathrm{We}\:\mathrm{know}\:\int\frac{{du}}{\alpha{u}^{\mathrm{2}} +\beta}=\frac{\mathrm{tan}^{−\mathrm{1}} \:\frac{{u}\sqrt{\alpha}}{\:\sqrt{\beta}}}{\:\sqrt{\alpha}\sqrt{\beta}}\:\mathrm{and}\:\sqrt{\left({a}\pm\mathrm{1}\right)^{\mathrm{2}} }=\mid{a}\pm\mathrm{1}\mid\right] \\ $$
Answered by Spillover last updated on 14/Dec/25
