Question Number 36892 by anik last updated on 06/Jun/18
![2. ∫[(√((1−x^2 )/(1+x^2 )))]dx=?](Q36892.png)
$$\mathrm{2}.\:\int\left[\sqrt{\left(\mathrm{1}−{x}^{\mathrm{2}} \right)/\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}\right]{dx}=? \\ $$
Commented by math khazana by abdo last updated on 11/Jun/18
$${let}\:{I}\:\:=\:\int\:\:\sqrt{\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{1}+{x}^{\mathrm{2}} }\:}\:{dx}\:{changement}\:{x}={sint}\:{give} \\ $$$${I}\:\:=\:\:\int\:\:\:\frac{{cost}}{\sqrt{\mathrm{2}{cos}^{\mathrm{2}} {t}\:−\mathrm{1}}}\:{costdt} \\ $$$$=\:\int\:\:\:\:\:\frac{{cos}^{\mathrm{2}} {t}}{\sqrt{\mathrm{2}\:{cos}^{\mathrm{2}} {t}\:−\mathrm{1}}}\:{dt}\:=\:\int\:\:\frac{\mathrm{1}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)\sqrt{\mathrm{2}\frac{\mathrm{1}}{\mathrm{1}+{tan}^{\mathrm{2}} {t}}−\mathrm{1}}}{dt} \\ $$$$=\:\int\:\:\:\:\:\:\:\:\frac{{dt}}{\left(\mathrm{1}+{tan}^{\mathrm{2}} {t}\right)\sqrt{\mathrm{2}−\mathrm{1}−{tan}^{\mathrm{2}} {t}}}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\:{dt} \\ $$$$=\:\int\:\:\:\:\:\:\:\frac{{dt}}{\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} {t}}\sqrt{\mathrm{1}−{tan}^{\mathrm{2}} {t}}}\:\:{changement} \\ $$$${tant}\:={u}\:\:{give} \\ $$$${I}\:=\:\:\int\:\:\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}+{u}^{\mathrm{2}} }\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}\:\:\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\:\int\:\:\:\:\:\frac{{du}}{\left(\mathrm{1}+{u}^{\mathrm{2}} \right)\sqrt{\mathrm{1}−{u}^{\mathrm{4}} }}\:\:....{be}\:{continued}.... \\ $$$$ \\ $$
Answered by anik last updated on 06/Jun/18
Commented by anik last updated on 06/Jun/18
$$\mathrm{Indrajit}\:\mathrm{Karmakar}\:\mathrm{have}\:\mathrm{solved}\:\mathrm{this}. \\ $$
Commented by MJS last updated on 06/Jun/18
$$\int\frac{\mathrm{cos}^{\mathrm{2}} \:\theta}{\sqrt{\mathrm{2}−\mathrm{cos}^{\mathrm{2}} \:\theta}}{d}\theta=\int\frac{{d}\theta}{\mathrm{sec}\:\theta\:\sqrt{\mathrm{2sec}^{\mathrm{2}} \:\theta\:−\mathrm{1}}} \\ $$$$\mathrm{so}\:\mathrm{something}\:\mathrm{went}\:\mathrm{wrong}... \\ $$