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Question Number 66740 by behi83417@gmail.com last updated on 19/Aug/19

$$\underset{\:\:\mathrm{0}} {\overset{\:\:\:\:\:\:\:\mathrm{1}} {\int}}\sqrt{\mathrm{1}−\boldsymbol{\mathrm{x}}+\boldsymbol{\mathrm{x}}^{\mathrm{2}} −\boldsymbol{\mathrm{x}}^{\mathrm{3}} \:}\:\:\boldsymbol{\mathrm{dx}}=? \\$$

Commented bymathmax by abdo last updated on 21/Aug/19

$${its}\:{only}\:{a}\:{try}\:{let}\:{I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} −{x}^{\mathrm{3}} }{dx}\:\Rightarrow \\$$ $${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}+{x}^{\mathrm{2}} \left(\mathrm{1}−{x}\right)}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left(\mathrm{1}−{x}\right)\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{dx} \\$$ $$=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{1}−{x}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{\mathrm{1}−{x}}={t}\:\Rightarrow\mathrm{1}−{x}={t}^{\mathrm{2}} \\$$ $${I}\:=−\int_{\mathrm{0}} ^{\mathrm{1}} {t}\sqrt{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }\left(−\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{\mathrm{2}} \sqrt{\mathrm{1}+\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\$$ $${changement}\:\mathrm{1}−{t}^{\mathrm{2}} ={tan}\theta\:{give}\:{t}^{\mathrm{2}} =\mathrm{1}−{tan}\theta\:\Rightarrow{t}=\sqrt{\mathrm{1}−{tan}\theta} \\$$ $${I}\:=\mathrm{2}\:\int_{\frac{\pi}{\mathrm{4}}} ^{\mathrm{0}} \left(\mathrm{1}−{tan}\theta\right)\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\frac{−\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\mathrm{2}\sqrt{\mathrm{1}−{tan}\theta}}\:{d}\theta \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}−{tan}\theta}\sqrt{\mathrm{1}+{tan}^{\mathrm{2}} \theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right){d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\mathrm{1}−{tan}\theta}\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {d}\theta \\$$ $$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{4}}} \sqrt{\frac{{cos}\theta−{sin}\theta}{{cos}\theta}}\frac{{d}\theta}{{cos}^{\mathrm{3}} \theta}\:\:\:\:{let}\:{try}\:{another}\:{way} \\$$ $${I}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\frac{\mathrm{1}−\left(−{x}\right)^{\mathrm{4}} }{\mathrm{1}+{x}}}{dx}\:=\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{\mathrm{1}+{x}^{\mathrm{4}} }}{\sqrt{\mathrm{1}+{x}}}\:{dx}\:\:\:\:\sqrt{\mathrm{1}+{x}}={t}\:\Rightarrow\mathrm{1}+{x}\:={t}^{\mathrm{2}} \:\Rightarrow \\$$ $${x}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow\:{I}\:=\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{\sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }}{{t}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\sqrt{\mathrm{1}+{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{1}}{dt} \\$$ $$=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \sqrt{{t}^{\mathrm{4}} −\mathrm{2}{t}^{\mathrm{2}} \:+\mathrm{2}}{dt}\:=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \sqrt{\mathrm{1}+\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\mathrm{2}} }{dt} \\$$ $${let}\:{do}\:{the}\:{changement}\:{t}^{\mathrm{2}} −\mathrm{1}\:={sh}\left({u}\right)\:\Rightarrow{t}^{\mathrm{2}} =\mathrm{1}+{sh}\left({u}\right)\:\Rightarrow{t}=\sqrt{\mathrm{1}+{sh}\left({u}\right)} \\$$ $${I}=\mathrm{2}\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} {ch}\left({u}\right)\:\:\frac{{chu}}{\mathrm{2}\sqrt{\mathrm{1}+{sh}\left({u}\right)}}\:{du}\:=\:\int_{\mathrm{1}} ^{\sqrt{\mathrm{2}}} \:\:\frac{{ch}^{\mathrm{2}} {u}}{\sqrt{\mathrm{1}+{shu}}}\:{du} \\$$ $${its}\:{seems}\:{that}\:{I}\:{is}\:{a}\:{elliptic}\:{integral}....{be}\:{continued}.... \\$$