Question Number 36940 by maxmathsup by imad last updated on 07/Jun/18
$${find}\:{f}\left({a}\right)=\:\int_{\mathrm{0}} ^{{a}} \:{arctan}\left(\sqrt{{a}^{\mathrm{2}} \:−{x}^{\mathrm{2}} }\right){dx} \\ $$
Commented by math khazana by abdo last updated on 08/Jun/18
![changement (√(a^2 −x^2 ))=t givea^2 −x^2 =t^2 ⇒ x^2 =a^2 −t^2 ⇒x=(√(a^2 −t^2 ))⇒dx = ((−2tdt)/(2(√(a^2 −t^2 )))) f(a) = ∫_0 ^a arctan(t) ((t dt)/(√(a^2 −t^2 ))) by parts f(a) =[−arctan(t)(√(a^2 −t^2 ))]_0 ^a −∫_0 ^a ((−(√(a^2 −t^2 )))/(1+t^2 )) dt = ∫_0 ^a ((√(a^2 −t^2 ))/(1+t^2 )) dt changement t= a sinα give f(a) = ∫_0 ^(π/2) ((∣a∣cosα)/(1+a^2 sin^2 α)) acosα dα =a∣a∣ ∫_0 ^(π/2) ((cos^2 α)/(1+a^2 sin^2 α))dα but ((cos^2 α)/(1+a^2 sin^2 α)) = ((1/(1+tan^2 α))/(1+a^2 (1−(1/(1+tan^2 α))))) = (1/((1+tan^2 α)(1+a^2 tan^2 α).(1/(1+tan^2 α)))) = (1/(1 +a^2 tan^2 α)) ⇒ f(a)=a∣a∣ ∫_0 ^(π/2) (dα/(1+a^2 tan^2 α)) =_(atan(α)=x) a∣a∣ ∫_0 ^(+∞) (1/(1+x^2 )) (1/(a( 1+(x^2 /a^2 ))))dx =∣a∣ ∫_0 ^∞ ((a^2 dx)/((1+x^2 )( x^2 +a^2 ))) case1 a>0 anda≠1⇒f(a) =(a^3 /2) ∫_(−∞) ^(+∞) (dx/((x^2 +1)(x^2 +a^2 ))) let ϕ(z) = (1/((z^2 +1)(z^2 +a^2 ))) ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ { Res(ϕ,i)+Res(ϕ,ia)} ϕ(z)= (1/((z−i)(z+i)(z−ia)(z+ia))) Res(ϕ,i)= (1/(2i(a^2 −1))) Res(ϕ^ ,ia) = (1/(2ia(1−a^2 ))) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ{ (1/(2i(a^2 −1))) +(1/(2ia(1−a^2 )))} =(π/(a^2 −1)){ 1−(1/a)}= ((π(a−1))/(a(a^2 −1))) =(π/(a(a+1))) f(a)= (a^3 /2) .(π/(a(a+1))) = ((πa^2 )/(2(a+1))) case 2 a<0 and a≠−1 we folow the same method...](Q37083.png)
$${changement}\:\sqrt{{a}^{\mathrm{2}} \:−{x}^{\mathrm{2}} }={t}\:{givea}^{\mathrm{2}} \:−{x}^{\mathrm{2}} \:={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${x}^{\mathrm{2}} ={a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} \:\Rightarrow{x}=\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }\Rightarrow{dx}\:=\:\frac{−\mathrm{2}{tdt}}{\mathrm{2}\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{{a}} \:{arctan}\left({t}\right)\:\frac{{t}\:{dt}}{\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}\:{by}\:{parts} \\ $$$${f}\left({a}\right)\:=\left[−{arctan}\left({t}\right)\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{{a}} \:\:−\int_{\mathrm{0}} ^{{a}} \:\:\frac{−\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt} \\ $$$$=\:\int_{\mathrm{0}} ^{{a}} \:\:\:\:\frac{\sqrt{{a}^{\mathrm{2}} \:−{t}^{\mathrm{2}} }}{\mathrm{1}+{t}^{\mathrm{2}} }\:{dt}\:{changement}\:{t}=\:{a}\:{sin}\alpha\:{give} \\ $$$${f}\left({a}\right)\:=\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\frac{\mid{a}\mid{cos}\alpha}{\mathrm{1}+{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha}\:{acos}\alpha\:{d}\alpha \\ $$$$={a}\mid{a}\mid\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{{cos}^{\mathrm{2}} \alpha}{\mathrm{1}+{a}^{\mathrm{2}} \:{sin}^{\mathrm{2}} \alpha}{d}\alpha \\ $$$${but}\:\frac{{cos}^{\mathrm{2}} \alpha}{\mathrm{1}+{a}^{\mathrm{2}} {sin}^{\mathrm{2}} \alpha}\:=\:\:\frac{\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}}{\mathrm{1}+{a}^{\mathrm{2}} \left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \alpha\right)\left(\mathrm{1}+{a}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \alpha\right).\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} \alpha}} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{1}\:+{a}^{\mathrm{2}} \:{tan}^{\mathrm{2}} \alpha}\:\Rightarrow\:{f}\left({a}\right)={a}\mid{a}\mid\:\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\:\frac{{d}\alpha}{\mathrm{1}+{a}^{\mathrm{2}} {tan}^{\mathrm{2}} \alpha} \\ $$$$=_{{atan}\left(\alpha\right)={x}} \:\:{a}\mid{a}\mid\:\int_{\mathrm{0}} ^{+\infty} \:\:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:\frac{\mathrm{1}}{{a}\left(\:\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\right)}{dx} \\ $$$$=\mid{a}\mid\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\:\:\frac{{a}^{\mathrm{2}} {dx}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\left(\:{x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)} \\ $$$${case}\mathrm{1}\:\:{a}>\mathrm{0}\:\:{anda}\neq\mathrm{1}\Rightarrow{f}\left({a}\right)\:=\frac{{a}^{\mathrm{3}} }{\mathrm{2}}\:\int_{−\infty} ^{+\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)\left({x}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)} \\ $$$${let}\:\varphi\left({z}\right)\:=\:\frac{\mathrm{1}}{\left({z}^{\mathrm{2}} \:+\mathrm{1}\right)\left({z}^{\mathrm{2}} \:+{a}^{\mathrm{2}} \right)}\: \\ $$$$\int_{−\infty} ^{+\infty} \:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\:\left\{\:{Res}\left(\varphi,{i}\right)+{Res}\left(\varphi,{ia}\right)\right\} \\ $$$$\varphi\left({z}\right)=\:\frac{\mathrm{1}}{\left({z}−{i}\right)\left({z}+{i}\right)\left({z}−{ia}\right)\left({z}+{ia}\right)} \\ $$$${Res}\left(\varphi,{i}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{i}\left({a}^{\mathrm{2}} −\mathrm{1}\right)} \\ $$$${Res}\left(\hat {\varphi},{ia}\right)\:=\:\frac{\mathrm{1}}{\mathrm{2}{ia}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\:\Rightarrow \\ $$$$\int_{−\infty} ^{+\infty} \:\:\varphi\left({z}\right){dz}\:=\mathrm{2}{i}\pi\left\{\:\:\frac{\mathrm{1}}{\mathrm{2}{i}\left({a}^{\mathrm{2}} \:−\mathrm{1}\right)}\:+\frac{\mathrm{1}}{\mathrm{2}{ia}\left(\mathrm{1}−{a}^{\mathrm{2}} \right)}\right\} \\ $$$$=\frac{\pi}{{a}^{\mathrm{2}} −\mathrm{1}}\left\{\:\mathrm{1}−\frac{\mathrm{1}}{{a}}\right\}=\:\frac{\pi\left({a}−\mathrm{1}\right)}{{a}\left({a}^{\mathrm{2}} −\mathrm{1}\right)}\:=\frac{\pi}{{a}\left({a}+\mathrm{1}\right)} \\ $$$${f}\left({a}\right)=\:\frac{{a}^{\mathrm{3}} }{\mathrm{2}}\:.\frac{\pi}{{a}\left({a}+\mathrm{1}\right)}\:=\:\frac{\pi{a}^{\mathrm{2}} }{\mathrm{2}\left({a}+\mathrm{1}\right)} \\ $$$${case}\:\mathrm{2}\:{a}<\mathrm{0}\:{and}\:{a}\neq−\mathrm{1}\:{we}\:{folow}\:{the}\:{same}\:{method}... \\ $$$$ \\ $$