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Question Number 51991 by maxmathsup by imad last updated on 01/Jan/19
find f(a) =∫ (dx/( (√(1+ax^2 ))+(√(1−ax^2 )))) with a>0
$${find}\:{f}\left({a}\right)\:=\int\:\:\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{ax}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{ax}^{\mathrm{2}} }}\:\:{with}\:{a}>\mathrm{0} \\ $$
Commented by Abdo msup. last updated on 05/Jan/19
changement x(√a)=t give f(a)=∫ (dt/( (√a)((√(1+t^2 ))+(√(1−t^2 ))))) =(1/( (√a))) ∫ (dt/( (√(1+t^2 ))+(√(1−t^2 )))) ⇒(√a)f(a)=∫ (((√(1+t^2 ))−(√(1−t^2 )))/(2t^2 ))dt =(1/2) ∫ ((√(1+t^2 ))/t^2 ) dt−(1/2) ∫ ((√(1−t^2 ))/t^2 ) dt but by parts ∫ ((√(1+t^2 ))/t^2 )dt =((−(√(1+t^2 )))/t) −∫ −(1/t) ((2t)/(2(√(1+t^2 ))))dt =−((√(1+t^2 ))/t) +ln(t+(√(1+t^2 ))) +c_1 also by psrts ∫ ((√(1−t^2 ))/t^2 ) dt =−((√(1−t^2 ))/t) −∫ −(1/t) ((−2t)/(2(√(1−t^2 )))) dt =−((√(1−t^2 ))/t) −arcsint +c_2 ⇒ f(a)=(1/(2(√a))){ −((√(1+t^2 ))/t) +ln(t+(√(1+t^2 )))+((√(1−t^2 ))/t) +arcsint } +C .
$${changement}\:{x}\sqrt{{a}}={t}\:{give} \\ $$$${f}\left({a}\right)=\int\:\:\:\:\:\:\:\:\frac{{dt}}{\:\sqrt{{a}}\left(\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\right)} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{{a}}}\:\int\:\:\:\:\:\frac{{dt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }+\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:\Rightarrow\sqrt{{a}}{f}\left({a}\right)=\int\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }−\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{\mathrm{2}{t}^{\mathrm{2}} }{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }\:{dt}−\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }\:{dt}\:\:{but}\:{by}\:{parts} \\ $$$$\int\:\:\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }{dt}\:=\frac{−\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}\:−\int\:−\frac{\mathrm{1}}{{t}}\:\:\frac{\mathrm{2}{t}}{\mathrm{2}\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{dt} \\ $$$$=−\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}\:+{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)\:+{c}_{\mathrm{1}} \:{also}\:{by}\:{psrts} \\ $$$$\int\:\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}^{\mathrm{2}} }\:{dt}\:=−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\:−\int\:\:−\frac{\mathrm{1}}{{t}}\:\frac{−\mathrm{2}{t}}{\mathrm{2}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}\:{dt} \\ $$$$=−\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\:−{arcsint}\:\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${f}\left({a}\right)=\frac{\mathrm{1}}{\mathrm{2}\sqrt{{a}}}\left\{\:−\frac{\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}{{t}}\:+{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\right)+\frac{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}{{t}}\right. \\ $$$$\left.+{arcsint}\:\right\}\:+{C}\:. \\ $$$$ \\ $$

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