Question Number 44304 by abdo.msup.com last updated on 26/Sep/18
$${find}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\frac{{dx}}{\mathrm{1}+{acos}^{\mathrm{2}} {x}} \\ $$$${a}\:{from}\:{R}. \\ $$
Commented by maxmathsup by imad last updated on 27/Sep/18
![f(a) = ∫_0 ^(π/4) (dx/(1+acos^2 x)) = ∫_0 ^(π/4) (dx/(1+a((1+cos(2x))/2))) = ∫_0 ^(π/4) ((2dx)/(2 +a +acos(2x))) =_(2x=t) ∫_0 ^(π/2) (dt/(2+a +acost)) changement tan((t/2))=u give f(a) = ∫_0 ^1 (1/(2+a +a ((1−u^2 )/(1+u^2 )))) ((2du)/(1+u^2 )) = ∫_0 ^1 ((2du)/(2+a +(2+a)u^2 +a −au^2 )) = ∫_0 ^1 ((2du)/(2u^2 +2+2a)) = ∫_0 ^1 (du/(u^2 +1+a)) case 1 1+a>0 ⇒f(a) =_(u=(√(1+a))α) ∫_0 ^(√(1+a)) (((√(1+a))dα)/((1+a)(1+α^2 ))) = (1/(√(1+a))) [ arctan(α)]_0 ^(√(1+a)) = ((arctan((√(1+a))))/(√(1+a))) case 2 1+a <0 ⇒f(a) = ∫_0 ^1 (du/(u^2 −((√(−1−a)))^2 )) =(1/(2(√(−1−a)))) ∫_0 ^1 { (1/(u−(√(−1−a)))) −(1/(u+(√(−1−a))))}du = (1/(2(√(−1−a)))) [ ln∣((u−(√(−1−a)))/(u+(√(−1−a))))∣]_0 ^1 = (1/(2(√(−1−a))))ln∣((1−(√(−1−a)))/(1+(√(−1−a))))∣ .](Q44359.png)
$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{dx}}{\mathrm{1}+{acos}^{\mathrm{2}} {x}}\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\frac{{dx}}{\mathrm{1}+{a}\frac{\mathrm{1}+{cos}\left(\mathrm{2}{x}\right)}{\mathrm{2}}} \\ $$$$=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \:\:\:\:\frac{\mathrm{2}{dx}}{\mathrm{2}\:+{a}\:+{acos}\left(\mathrm{2}{x}\right)}\:=_{\mathrm{2}{x}={t}} \:\:\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\frac{{dt}}{\mathrm{2}+{a}\:+{acost}}\:{changement}\:{tan}\left(\frac{{t}}{\mathrm{2}}\right)={u} \\ $$$${give}\:{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{2}+{a}\:+{a}\:\frac{\mathrm{1}−{u}^{\mathrm{2}} }{\mathrm{1}+{u}^{\mathrm{2}} }}\:\frac{\mathrm{2}{du}}{\mathrm{1}+{u}^{\mathrm{2}} }\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}+{a}\:+\left(\mathrm{2}+{a}\right){u}^{\mathrm{2}} \:+{a}\:−{au}^{\mathrm{2}} } \\ $$$$=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{2}+\mathrm{2}{a}}\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\mathrm{1}+{a}} \\ $$$${case}\:\mathrm{1}\:\:\:\:\mathrm{1}+{a}>\mathrm{0}\:\:\Rightarrow{f}\left({a}\right)\:=_{{u}=\sqrt{\mathrm{1}+{a}}\alpha} \:\:\int_{\mathrm{0}} ^{\sqrt{\mathrm{1}+{a}}} \:\:\:\:\frac{\sqrt{\mathrm{1}+{a}}{d}\alpha}{\left(\mathrm{1}+{a}\right)\left(\mathrm{1}+\alpha^{\mathrm{2}} \right)} \\ $$$$=\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{a}}}\:\left[\:{arctan}\left(\alpha\right)\right]_{\mathrm{0}} ^{\sqrt{\mathrm{1}+{a}}} =\:\frac{{arctan}\left(\sqrt{\mathrm{1}+{a}}\right)}{\sqrt{\mathrm{1}+{a}}} \\ $$$${case}\:\mathrm{2}\:\:\mathrm{1}+{a}\:<\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\:\:\frac{{du}}{{u}^{\mathrm{2}} −\left(\sqrt{−\mathrm{1}−{a}}\right)^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\mathrm{1}−{a}}}\:\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\:\:\frac{\mathrm{1}}{{u}−\sqrt{−\mathrm{1}−{a}}}\:−\frac{\mathrm{1}}{{u}+\sqrt{−\mathrm{1}−{a}}}\right\}{du} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\mathrm{1}−{a}}}\:\left[\:{ln}\mid\frac{{u}−\sqrt{−\mathrm{1}−{a}}}{{u}+\sqrt{−\mathrm{1}−{a}}}\mid\right]_{\mathrm{0}} ^{\mathrm{1}} =\:\frac{\mathrm{1}}{\mathrm{2}\sqrt{−\mathrm{1}−{a}}}{ln}\mid\frac{\mathrm{1}−\sqrt{−\mathrm{1}−{a}}}{\mathrm{1}+\sqrt{−\mathrm{1}−{a}}}\mid\:. \\ $$
Commented by maxmathsup by imad last updated on 27/Sep/18
$${if}\:{a}=−\mathrm{1}\:\:{f}\left({a}\right)\:{is}\:{divergent}\:. \\ $$
Answered by Smail last updated on 27/Sep/18
![f(a)=∫_0 ^(π/4) (dx/(1+acos^2 x)) t=tanx⇒dx=(dt/(1+t^2 )) f(a)=∫_0 ^1 (dt/((1+t^2 )(1+(a/(1+t^2 ))))) =∫_0 ^1 (dt/(1+t^2 +a))=∫_0 ^1 (dt/((1+a)((t^2 /(a+1))+1))) =(1/(a+1))∫_0 ^1 (dt/(((t/(√(a+1))))^2 +1)) u=(t/(√(a+1)))⇒dt=(√(a+1))du f(a)=((√(a+1))/(a+1))∫_0 ^(1/(√(a+1))) (du/(u^2 +1)) =((√(a+1))/(a+1))[tan^(−1) (u)]_0 ^(1/(√(a+1))) =((√(a+1))/(a+1))tan^(−1) (((√(a+1))/(a+1)))](Q44327.png)
$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{4}} \frac{{dx}}{\mathrm{1}+{acos}^{\mathrm{2}} {x}} \\ $$$${t}={tanx}\Rightarrow{dx}=\frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} } \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(\mathrm{1}+\frac{{a}}{\mathrm{1}+{t}^{\mathrm{2}} }\right)} \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\mathrm{1}+{t}^{\mathrm{2}} +{a}}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left(\mathrm{1}+{a}\right)\left(\frac{{t}^{\mathrm{2}} }{{a}+\mathrm{1}}+\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{{a}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dt}}{\left(\frac{{t}}{\sqrt{{a}+\mathrm{1}}}\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$${u}=\frac{{t}}{\sqrt{{a}+\mathrm{1}}}\Rightarrow{dt}=\sqrt{{a}+\mathrm{1}}{du} \\ $$$${f}\left({a}\right)=\frac{\sqrt{{a}+\mathrm{1}}}{{a}+\mathrm{1}}\int_{\mathrm{0}} ^{\mathrm{1}/\sqrt{{a}+\mathrm{1}}} \frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\frac{\sqrt{{a}+\mathrm{1}}}{{a}+\mathrm{1}}\left[{tan}^{−\mathrm{1}} \left({u}\right)\right]_{\mathrm{0}} ^{\mathrm{1}/\sqrt{{a}+\mathrm{1}}} \\ $$$$=\frac{\sqrt{{a}+\mathrm{1}}}{{a}+\mathrm{1}}{tan}^{−\mathrm{1}} \left(\frac{\sqrt{{a}+\mathrm{1}}}{{a}+\mathrm{1}}\right) \\ $$