Question Number 65355 by mathmax by abdo last updated on 28/Jul/19
$${find}\:\int\:\:\:\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}} \\ $$
Commented by Prithwish sen last updated on 29/Jul/19
$$\int\frac{\mathrm{dx}}{\:\sqrt{\left\{\left(\mathrm{x}+\mathrm{2}\right)^{\mathrm{2}} −\mathrm{1}\right\}\left(\mathrm{x}+\mathrm{2}\right)}}\:\:\:\mathrm{putting}\:\left(\mathrm{x}+\mathrm{2}\right)=\mathrm{a} \\ $$$$\:=\int\frac{\mathrm{da}}{\:\sqrt{\mathrm{a}\left(\mathrm{a}^{\mathrm{2}} −\mathrm{1}\right)}}\:\:\mathrm{again}\:\mathrm{putting}\:\mathrm{a}=\mathrm{sec}\theta \\ $$$$=\:\int\sqrt{\mathrm{sec}\theta}\:\mathrm{d}\theta\:\:???? \\ $$
Answered by ajfour last updated on 29/Jul/19
$${x}+\mathrm{2}={t} \\ $$$$\int\frac{{dx}}{\:\sqrt{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}} \\ $$$${let}\:{t}=\mathrm{sec}\:^{\mathrm{2}} \theta\:\Rightarrow\:\:{dt}=\mathrm{2sec}\:^{\mathrm{2}} \theta\mathrm{tan}\:\theta{d}\theta \\ $$$$\int\frac{\mathrm{2sec}\:\theta\mathrm{tan}\:\theta{d}\theta}{\:\sqrt{\mathrm{sec}\:^{\mathrm{4}} \theta−\mathrm{1}}} \\ $$$${let}\:\:\mathrm{sec}\:\theta={z} \\ $$$$\int\frac{\mathrm{2}{dz}}{\:\sqrt{{z}^{\mathrm{4}} −\mathrm{1}}}=\:? \\ $$
Answered by MJS last updated on 29/Jul/19
![∫(dx/( (√((x+1)(x+2)(x+3)))))= [t=arccosh (x+2) → dx=dt(√((x+1)(x+3)))] =∫(dt/( (√(cosh t))))= [u=(t/2)i → dt=−2idu] =−2i∫(du/( (√(cos 2u))))=−2i∫(du/( (√(1−2sin^2 u))))= this is an incomplete elliptic integral =−2iF(u∣2) ...](https://www.tinkutara.com/question/Q65422.png)
$$\int\frac{{dx}}{\:\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccosh}\:\left({x}+\mathrm{2}\right)\:\rightarrow\:{dx}={dt}\sqrt{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{3}\right)}\right] \\ $$$$=\int\frac{{dt}}{\:\sqrt{\mathrm{cosh}\:{t}}}= \\ $$$$\:\:\:\:\:\left[{u}=\frac{{t}}{\mathrm{2}}\mathrm{i}\:\rightarrow\:{dt}=−\mathrm{2i}{du}\right] \\ $$$$=−\mathrm{2i}\int\frac{{du}}{\:\sqrt{\mathrm{cos}\:\mathrm{2}{u}}}=−\mathrm{2i}\int\frac{{du}}{\:\sqrt{\mathrm{1}−\mathrm{2sin}^{\mathrm{2}} \:{u}}}= \\ $$$$\mathrm{this}\:\mathrm{is}\:\mathrm{an}\:\mathrm{incomplete}\:\mathrm{elliptic}\:\mathrm{integral} \\ $$$$=−\mathrm{2i}{F}\left({u}\mid\mathrm{2}\right) \\ $$$$… \\ $$