Question Number 82330 by Power last updated on 20/Feb/20
Commented by mathmax by abdo last updated on 20/Feb/20
$${let}\:{I}\:=\int{x}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}\:\:{changement}\:{x}={ash}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:{a}^{\mathrm{2}} {sh}^{\mathrm{2}} \left({t}\right)\mid{a}\mid{ch}\left({t}\right){ach}\left({t}\right){dt}\:={a}^{\mathrm{3}} \mid{a}\mid\int{sh}^{\mathrm{2}} \left({t}\right){ch}^{\mathrm{2}} \left({t}\right){dt} \\ $$$$={a}^{\mathrm{3}} \mid{a}\mid\:\int\left(\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left({ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right)\left({ch}\left(\mathrm{2}{t}\right)+\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left({ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{1}\right){dt}\:\:=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left(\frac{\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)}{\mathrm{2}}−\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}\int\:\left({ch}\left(\mathrm{4}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}×\mathrm{4}}{sh}\left(\mathrm{4}{t}\right)−\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}{t}\:+{C} \\ $$$${sh}\left(\mathrm{4}{t}\right)\:=\frac{{e}^{\mathrm{4}{t}} −{e}^{−\mathrm{4}{t}} }{\mathrm{2}}\:\:{and}\:{t}={argsh}\left(\frac{{x}}{{a}}\right)\:={ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:\Rightarrow \\ $$$${sh}\left(\mathrm{4}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right\}\:\Rightarrow \\ $$$${I}\:=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{8}×\mathrm{32}}\left\{\:\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right\} \\ $$$$−\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}{ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:+{C}\:. \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{sir}\:\:\mathrm{x}=\mathrm{a}\:\mathrm{tanu}\:\:\:\mathrm{solution} \\ $$
Commented by john santu last updated on 20/Feb/20
$$\int\:{x}\:\left({x}\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right)\:{dx}\:=\:{I} \\ $$$${u}\:=\:{x}\:\Rightarrow{du}\:=\:{dx} \\ $$$${v}\:=\:\int\:{x}\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{d}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{x}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\int\:\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx} \\ $$$${J}\:=\:\int\:\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dx} \\ $$$${let}\:{x}\:=\:{a}\:\mathrm{tan}\:\theta\:\Rightarrow{dx}\:=\:{a}\mathrm{sec}\:\theta\:{d}\theta \\ $$$${J}\:=\:\int{a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{4}} \theta\:{d}\theta\: \\ $$$${J}\:=\:{a}^{\mathrm{2}} \int\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right)\:{d}\left(\mathrm{tan}\:\theta\right) \\ $$$${J}\:=\:{a}^{\mathrm{2}} \left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \theta\right)+{c} \\ $$$${J}\:=\:{a}^{\mathrm{2}} \left(\frac{{x}}{{a}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}{a}^{\mathrm{3}} }\right)+{c}\:=\:{ax}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\frac{{x}^{\mathrm{3}} }{{a}}+{c} \\ $$$$\therefore\:{I}\:−{J}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:−\:{ax}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}{a}}\:+\:{c} \\ $$
Commented by Power last updated on 20/Feb/20
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$