Question Number 41913 by math khazana by abdo last updated on 15/Aug/18
$${let}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\:\frac{{x}}{\mathrm{1}+{acosx}}{dx} \\ $$$$\left.\mathrm{1}\right)\:{find}\:\:{f}\left({a}\right)\: \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}}{\mathrm{1}+\mathrm{2}{cosx}}{dx}\:{and}\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}}{\mathrm{1}−\mathrm{2}{cosx}}{dx} \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\frac{{xcosx}}{\left(\mathrm{1}+{acosx}\right)^{\mathrm{2}} }{dx} \\ $$$$\left.\mathrm{4}\right)\:{find}\:{the}\:{value}\:{of}\:\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{xcosx}}{\left(\mathrm{1}+\mathrm{2}{cosx}\right)^{\mathrm{2}} }{dx}\:. \\ $$
Answered by maxmathsup by imad last updated on 16/Aug/18
![1) we have f(a) = ∫_0 ^π (x/(1+a((2tan((x/2)))/(1+tan^2 ((x/2)))))) dx changement tan((x/2)) =t give f(a) =∫_0 ^∞ ((2arctan(t))/(1+((2at)/(1+t^2 )))) ((2dt)/(1+t^2 )) =∫_0 ^∞ ((4arctan(t))/(1+t^2 +2at))dt =∫_0 ^∞ ((4arctan(t))/(t^2 +2at +1)) dt =4 ∫_0 ^∞ ((arctan(t))/(t^2 +2at +1))dt =w(1) with w(α) =∫_0 ^∞ ((arctan(αt))/(t^2 +2at +1)) dt we have w^′ (α) =∫_0 ^∞ (t/((1+α^2 t^2 )(t^2 +2at +1))) dt =_(αt =u) ∫_0 ^∞ (u/(α(1+u^2 )((u^2 /α^2 ) +2a(u/α)+1)))(du/α) = ∫_0 ^∞ ((udu)/(α^2 (1+u^2 )(((u^2 +2αau +α^2 )/α^2 )))) =∫_0 ^∞ ((udu)/((u^2 +1)(u^2 +2αau +α^2 ))) let decompose F(u) = (u/((u^2 +1)(u^2 +2αau +α^2 ))) (let take α>0) roots of u^2 +2αau +α^2 Δ^′ =α^2 a^2 −α^2 =α^2 (a^2 −1) case1) ∣a∣>1 ⇒ u_1 =−αa +∣α∣(√(a^2 −1)) and u_2 =−αa −∣α∣(√(a^2 −1)) F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +((cu +d)/(u^2 +1)) =(u/((u−u_1 )(u−u_2 )(u^2 +1))) a = (u_1 /((u_1 −u_2 )(u_1 ^2 +1))) =((∣α∣(√(a^2 −1))−αa)/(2∣α∣(√(a^2 −1))((∣α∣(√(a^2 −1))−αa)^2 +1))) =((α((√(a^2 −1))−a))/(2α(√(a^2 −1))(α^2 ((√(a^2 −1))−a)^2 +1))) =(((√(a^2 −1))−a)/(2(√(a^2 −1)){α^2 ((√(a^2 −1))−a)^2 +1})) b = (u_2 /((u_2 −u_1 )(u_2 ^2 +1))) = ((−α(a+(√(a^2 −1))))/(−2α(√(a^2 −1)){ α^2 (a+(√(a^2 −1)))^2 +1})) lim_(u→+∞) u F(u) =0 =a+b +c ⇒ c =−a−b ⇒ F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u +d)/(u^2 +1)) F(0) =0 =−(a/u_1 ) −(b/u_2 ) +d ⇒ d =(a/u_1 ) +(b/u_2 ) ⇒ F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) +(((−a−b)u +(a/u_1 )+(b/u_2 ))/(u^2 +1)) ⇒ ∫_0 ^∞ F(u)du =[aln∣u−u_1 ∣ +bln∣u−u_2 ∣]_0 ^(+∞) −[((a+b)/2)ln(u^2 +1)]_0 ^(+∞) +((a/u_1 ) +(b/u_2 ))[ arctan(u)]_0 ^(+∞) =[ ln(∣u−u_1 ∣^a ∣u−u_2 ∣^b )−ln((u^2 +1)^((a+b)/2) )]_0 ^(+∞) +(π/2){(a/u_1 ) +(b/u_2 )} ...be continued...](https://www.tinkutara.com/question/Q42018.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\pi} \:\:\:\frac{{x}}{\mathrm{1}+{a}\frac{\mathrm{2}{tan}\left(\frac{{x}}{\mathrm{2}}\right)}{\mathrm{1}+{tan}^{\mathrm{2}} \left(\frac{{x}}{\mathrm{2}}\right)}}\:{dx}\:\:{changement}\:{tan}\left(\frac{{x}}{\mathrm{2}}\right)\:={t}\:{give} \\ $$$${f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{2}{arctan}\left({t}\right)}{\mathrm{1}+\frac{\mathrm{2}{at}}{\mathrm{1}+{t}^{\mathrm{2}} }}\:\frac{\mathrm{2}{dt}}{\mathrm{1}+{t}^{\mathrm{2}} }\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{arctan}\left({t}\right)}{\mathrm{1}+{t}^{\mathrm{2}} \:+\mathrm{2}{at}}{dt}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{4}{arctan}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}}\:{dt} \\ $$$$=\mathrm{4}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{arctan}\left({t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}}{dt}\:={w}\left(\mathrm{1}\right)\:{with}\:{w}\left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{arctan}\left(\alpha{t}\right)}{{t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}}\:{dt}\:{we}\:{have} \\ $$$${w}^{'} \left(\alpha\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{t}}{\left(\mathrm{1}+\alpha^{\mathrm{2}} {t}^{\mathrm{2}} \right)\left({t}^{\mathrm{2}} \:+\mathrm{2}{at}\:+\mathrm{1}\right)}\:{dt}\:\:=_{\alpha{t}\:={u}} \:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{u}}{\alpha\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\frac{{u}^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\:+\mathrm{2}{a}\frac{{u}}{\alpha}+\mathrm{1}\right)}\frac{{du}}{\alpha} \\ $$$$=\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\:\frac{{udu}}{\alpha^{\mathrm{2}} \left(\mathrm{1}+{u}^{\mathrm{2}} \right)\left(\frac{{u}^{\mathrm{2}} \:+\mathrm{2}\alpha{au}\:+\alpha^{\mathrm{2}} }{\alpha^{\mathrm{2}} }\right)}\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{udu}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{2}\alpha{au}\:+\alpha^{\mathrm{2}} \right)} \\ $$$${let}\:{decompose}\:{F}\left({u}\right)\:=\:\frac{{u}}{\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\left({u}^{\mathrm{2}} \:+\mathrm{2}\alpha{au}\:+\alpha^{\mathrm{2}} \right)}\:\left({let}\:{take}\:\alpha>\mathrm{0}\right) \\ $$$${roots}\:{of}\:\:{u}^{\mathrm{2}} \:+\mathrm{2}\alpha{au}\:+\alpha^{\mathrm{2}} \\ $$$$\Delta^{'} \:=\alpha^{\mathrm{2}} {a}^{\mathrm{2}} \:−\alpha^{\mathrm{2}} \:=\alpha^{\mathrm{2}} \left({a}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$$\left.{case}\mathrm{1}\right)\:\mid{a}\mid>\mathrm{1}\:\:\Rightarrow\:{u}_{\mathrm{1}} =−\alpha{a}\:+\mid\alpha\mid\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}}\:\:\:{and}\:\:{u}_{\mathrm{2}} =−\alpha{a}\:−\mid\alpha\mid\sqrt{{a}^{\mathrm{2}} \:−\mathrm{1}} \\ $$$${F}\left({u}\right)\:=\:\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:\:+\frac{{cu}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\:=\frac{{u}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)} \\ $$$${a}\:=\:\:\frac{{u}_{\mathrm{1}} }{\left({u}_{\mathrm{1}} −{u}_{\mathrm{2}} \right)\left({u}_{\mathrm{1}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\frac{\mid\alpha\mid\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\alpha{a}}{\mathrm{2}\mid\alpha\mid\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\left(\left(\mid\alpha\mid\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−\alpha{a}\right)^{\mathrm{2}} +\mathrm{1}\right)}\:=\frac{\alpha\left(\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−{a}\right)}{\mathrm{2}\alpha\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\left(\alpha^{\mathrm{2}} \left(\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−{a}\right)^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$=\frac{\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−{a}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\left\{\alpha^{\mathrm{2}} \left(\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}−{a}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}} \\ $$$${b}\:=\:\frac{{u}_{\mathrm{2}} }{\left({u}_{\mathrm{2}} −{u}_{\mathrm{1}} \right)\left({u}_{\mathrm{2}} ^{\mathrm{2}} \:+\mathrm{1}\right)}\:=\:\frac{−\alpha\left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\right)}{−\mathrm{2}\alpha\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\left\{\:\alpha^{\mathrm{2}} \left({a}+\sqrt{{a}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\mathrm{1}\right\}} \\ $$$${lim}_{{u}\rightarrow+\infty} {u}\:{F}\left({u}\right)\:=\mathrm{0}\:={a}+{b}\:+{c}\:\Rightarrow\:{c}\:=−{a}−{b}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{\left(−{a}−{b}\right){u}\:+{d}}{{u}^{\mathrm{2}} \:+\mathrm{1}} \\ $$$${F}\left(\mathrm{0}\right)\:=\mathrm{0}\:=−\frac{{a}}{{u}_{\mathrm{1}} }\:−\frac{{b}}{{u}_{\mathrm{2}} }\:+{d}\:\Rightarrow\:{d}\:=\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:+\frac{\left(−{a}−{b}\right){u}\:\:+\frac{{a}}{{u}_{\mathrm{1}} }+\frac{{b}}{{u}_{\mathrm{2}} }}{{u}^{\mathrm{2}} \:+\mathrm{1}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:{F}\left({u}\right){du}\:\:=\left[{aln}\mid{u}−{u}_{\mathrm{1}} \mid\:+{bln}\mid{u}−{u}_{\mathrm{2}} \mid\right]_{\mathrm{0}} ^{+\infty} \:\:−\left[\frac{{a}+{b}}{\mathrm{2}}{ln}\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$+\left(\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\right)\left[\:{arctan}\left({u}\right)\right]_{\mathrm{0}} ^{+\infty} \\ $$$$=\left[\:{ln}\left(\mid{u}−{u}_{\mathrm{1}} \mid^{{a}} \mid{u}−{u}_{\mathrm{2}} \mid^{{b}} \right)−{ln}\left(\left({u}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{{a}+{b}}{\mathrm{2}}} \right)\right]_{\mathrm{0}} ^{+\infty} \:+\frac{\pi}{\mathrm{2}}\left\{\frac{{a}}{{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}_{\mathrm{2}} }\right\}\:\:…{be}\:{continued}… \\ $$$$ \\ $$