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Question Number 35687 by prof Abdo imad last updated on 22/May/18

calculate f(a)=∫_0 ^π (dx/(1−a cosx)) a from R . 2) application calculate ∫_0 ^π (dx/(1−2cosx))

$${calculate}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\frac{{dx}}{\mathrm{1}−{a}\:{cosx}}\:\:{a}\:{from}\:{R}\:. \\ $$$$\left.\mathrm{2}\right)\:{application}\:\:{calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{cosx}} \\ $$

Commented by prof Abdo imad last updated on 22/May/18

vhangement tan((x/2))=t give f(a) = ∫_0 ^∞ (1/(1−a ((1−t^2 )/(1+t^2 )))) ((2dt)/(1+t^2 )) = ∫_0 ^∞ ((2dt)/(1+t^2 −a(1−t^2 ))) = ∫_0 ^∞ ((2dt)/(1−a +(1+a)t^2 )) = (1/(1−a))∫_0 ^∞ ((2dt)/(1+((1+a)/(1−a))t^2 )) case 1 ((1+a)/(1−a))>0 and a≠1 ⇒ (((1+a)(1−a))/((1−a)^2 ))>0 ⇒ 1−a^2 >0 ⇒ ∣a∣<1 we get f(a) = (2/(1−a))∫_0 ^∞ (dt/(1+((√((1+a)/(1−a)))t)^2 )) =_(u= (√((1+a)/(1−a)))t) (2/(1−a))∫_0 ^∞ (1/(1+u^2 )) ((√(1−a))/(√(1+a))) du =(2/((√(1−a))(√(1+a)))) .(π/2) = (π/(√(1−a^2 ))) case2 ((1+a)/(1−a))<0 ⇒ ((1+a)/(a−1))>0 and f(a) = (1/(1−a))∫_0 ^∞ ((2dt)/(1 −((a+1)/(a−1))t^2 )) =_((√((a+1)/(a−1))) t =u) (2/(1−a)) ∫_0 ^∞ (1/(1 −u^2 )) ((√(a−1))/(√(a+1))) du = −(2/(√(a^2 −1))) ∫_0 ^∞ (du/(1−u^2 )) = (1/(√(a^2 −1))) ∫_0 ^∞ { (1/(u−1)) −(1/(u+1))}du = (1/(√(a^2 −1))) [ln∣((u−1)/(u+1))∣]_0 ^(+∞) =0 .

$${vhangement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)={t}\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}−{a}\:\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} \:−{a}\left(\mathrm{1}−{t}^{\mathrm{2}} \right)}\:=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}−{a}\:\:+\left(\mathrm{1}+{a}\right){t}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}+\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}{t}^{\mathrm{2}} } \\ $$$${case}\:\mathrm{1}\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}>\mathrm{0}\:\:{and}\:{a}\neq\mathrm{1}\:\:\Rightarrow\:\frac{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}−{a}\right)}{\left(\mathrm{1}−{a}\right)^{\mathrm{2}} }>\mathrm{0}\:\Rightarrow \\ $$$$\:\mathrm{1}−{a}^{\mathrm{2}} >\mathrm{0}\:\Rightarrow\:\mid{a}\mid<\mathrm{1}\:\:\:{we}\:{get} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dt}}{\mathrm{1}+\left(\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{t}\right)^{\mathrm{2}} } \\ $$$$=_{{u}=\:\sqrt{\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}}{t}} \:\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{\mathrm{2}} }\:\frac{\sqrt{\mathrm{1}−{a}}}{\sqrt{\mathrm{1}+{a}}}\:{du} \\ $$$$=\frac{\mathrm{2}}{\sqrt{\mathrm{1}−{a}}\sqrt{\mathrm{1}+{a}}}\:.\frac{\pi}{\mathrm{2}}\:=\:\:\frac{\pi}{\sqrt{\mathrm{1}−{a}^{\mathrm{2}} }} \\ $$$${case}\mathrm{2}\:\:\frac{\mathrm{1}+{a}}{\mathrm{1}−{a}}<\mathrm{0}\:\:\Rightarrow\:\frac{\mathrm{1}+{a}}{{a}−\mathrm{1}}>\mathrm{0}\:{and} \\ $$$${f}\left({a}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}−{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{dt}}{\mathrm{1}\:\:−\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}{t}^{\mathrm{2}} } \\ $$$$=_{\sqrt{\frac{{a}+\mathrm{1}}{{a}−\mathrm{1}}}\:{t}\:={u}} \:\:\frac{\mathrm{2}}{\mathrm{1}−{a}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}\:−{u}^{\mathrm{2}} }\:\frac{\sqrt{{a}−\mathrm{1}}}{\sqrt{{a}+\mathrm{1}}}\:{du} \\ $$$$=\:−\frac{\mathrm{2}}{\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{du}}{\mathrm{1}−{u}^{\mathrm{2}} } \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}}\:\int_{\mathrm{0}} ^{\infty} \left\{\:\frac{\mathrm{1}}{{u}−\mathrm{1}}\:−\frac{\mathrm{1}}{{u}+\mathrm{1}}\right\}{du} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}}\:\left[{ln}\mid\frac{{u}−\mathrm{1}}{{u}+\mathrm{1}}\mid\right]_{\mathrm{0}} ^{+\infty} \:=\mathrm{0}\:. \\ $$

Commented by prof Abdo imad last updated on 22/May/18

2)we have a =2 and ∣a∣>1 ⇒ ∫_0 ^π (dx/(1−2cosx)) =0

$$\left.\mathrm{2}\right){we}\:{have}\:{a}\:=\mathrm{2}\:{and}\:\mid{a}\mid>\mathrm{1}\:\Rightarrow\: \\ $$$$\int_{\mathrm{0}} ^{\pi} \:\:\:\:\:\:\frac{{dx}}{\mathrm{1}−\mathrm{2}{cosx}}\:=\mathrm{0} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 22/May/18

∫_0 ^Π (dx/(1−acosx)) =(1/a)∫_0 ^Π (dx/((1/a)−cosx)) let k=1/a =k∫_0 ^Π (dx/(k−((1−tan^2 ((x/2)))/(1+tan^2 ((x/2)))))) =k∫_0 ^Π (((1+tan^2 (x/2)))/(k+ktan^2 (x/2)−1+tan^2 (x/2)))dx t=tan(x/2) dt=sec^2 (x/2)×(1/2)dx =k∫_0 ^∞ ((2dt)/((k−1)+(k+1)t^2 )) =2k∫_0 ^∞ (dt/((k+1)(((k−1)/(k+1))+t^2 ))) =((2k)/(k+1))∫_0 ^∞ (dt/(((k−1)/(k+1))+t^2 )) =((2k)/(k+1))×(1/(√((k−1)/(k+1))))×∣tan^(−1) ((t/(√((k−1)/(k+1)))) )∣_0 ^∞ =((2k)/(k+1))×(((√(k+1)) )/(√(k−1)))×((Π/2)) =(2/(1+(1/k)))×((√(1+(1/k)))/(√(1−(1/k) )))×((Π/2)) =(2/(1+a))×((√(1+a))/(√(1−a)))×((Π/2))

$$\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{\mathrm{1}−{acosx}} \\ $$$$=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{\frac{\mathrm{1}}{{a}}−{cosx}} \\ $$$${let}\:{k}=\mathrm{1}/{a} \\ $$$$={k}\int_{\mathrm{0}} ^{\Pi} \frac{{dx}}{{k}−\frac{\mathrm{1}−{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}} \\ $$$$={k}\int_{\mathrm{0}} ^{\Pi} \frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}\right)}{{k}+{ktan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}−\mathrm{1}+{tan}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}}{dx} \\ $$$${t}={tan}\frac{{x}}{\mathrm{2}}\:{dt}={sec}^{\mathrm{2}} \frac{{x}}{\mathrm{2}}×\frac{\mathrm{1}}{\mathrm{2}}{dx} \\ $$$$={k}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{2}{dt}}{\left({k}−\mathrm{1}\right)+\left({k}+\mathrm{1}\right){t}^{\mathrm{2}} } \\ $$$$=\mathrm{2}{k}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\left({k}+\mathrm{1}\right)\left(\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}+{t}^{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}\int_{\mathrm{0}} ^{\infty} \frac{{dt}}{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}+{t}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}×\frac{\mathrm{1}}{\sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}}}×\mid{tan}^{−\mathrm{1}} \left(\frac{{t}}{\sqrt{\frac{{k}−\mathrm{1}}{{k}+\mathrm{1}}}}\:\:\right)\mid_{\mathrm{0}} ^{\infty} \\ $$$$=\frac{\mathrm{2}{k}}{{k}+\mathrm{1}}×\frac{\sqrt{{k}+\mathrm{1}}\:}{\sqrt{{k}−\mathrm{1}}}×\left(\frac{\Pi}{\mathrm{2}}\right) \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+\frac{\mathrm{1}}{{k}}}×\frac{\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{k}}}}{\sqrt{\mathrm{1}−\frac{\mathrm{1}}{{k}}\:}}×\left(\frac{\Pi}{\mathrm{2}}\right)\:\: \\ $$$$=\frac{\mathrm{2}}{\mathrm{1}+{a}}×\frac{\sqrt{\mathrm{1}+{a}}}{\sqrt{\mathrm{1}−{a}}}×\left(\frac{\Pi}{\mathrm{2}}\right)\: \\ $$$$ \\ $$

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