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Question Number 82330 by Power last updated on 20/Feb/20

Commented by mathmax by abdo last updated on 20/Feb/20

let I =∫x^2 (√(a^2 +x^2 ))dx changement x=ash(t) give I =∫ a^2 sh^2 (t)∣a∣ch(t)ach(t)dt =a^3 ∣a∣∫sh^2 (t)ch^2 (t)dt =a^3 ∣a∣ ∫( ((ch(2t)−1)/2))^2 ×(((1+ch(2t))/2))^2 dt =((a^3 ∣a∣)/(16)) ∫(ch(2t)−1)(ch(2t)+1)dt =((a^3 ∣a∣)/(16)) ∫(ch^2 (2t)−1)dt =((a^3 ∣a∣)/(16)) ∫(((1+ch(4t))/2)−1)dt =((a^3 ∣a∣)/(32))∫ (ch(4t)−1)dt =((a^3 ∣a∣)/(32×4))sh(4t)−((a^3 ∣a∣)/(32))t +C sh(4t) =((e^(4t) −e^(−4t) )/2) and t=argsh((x/a)) =ln((x/a)+(√(1+(x^2 /a^2 )))) ⇒ sh(4t) =(1/2){((x/a)+(√(1+(x^2 /a^2 ))))^4 −((x/a)+(√(1+(x^2 /a^2 ))))^(−4) } ⇒ I =((a^3 ∣a∣)/(8×32)){ ((x/a)+(√(1+(x^2 /a^2 ))))^4 −((x/a)+(√(1+(x^2 /a^2 ))))^(−4) } −((a^3 ∣a∣)/(32))ln((x/a)+(√(1+(x^2 /a^2 )))) +C .

$${let}\:{I}\:=\int{x}^{\mathrm{2}} \sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }{dx}\:\:{changement}\:{x}={ash}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:{a}^{\mathrm{2}} {sh}^{\mathrm{2}} \left({t}\right)\mid{a}\mid{ch}\left({t}\right){ach}\left({t}\right){dt}\:={a}^{\mathrm{3}} \mid{a}\mid\int{sh}^{\mathrm{2}} \left({t}\right){ch}^{\mathrm{2}} \left({t}\right){dt} \\ $$$$={a}^{\mathrm{3}} \mid{a}\mid\:\int\left(\:\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} ×\left(\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left({ch}\left(\mathrm{2}{t}\right)−\mathrm{1}\right)\left({ch}\left(\mathrm{2}{t}\right)+\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left({ch}^{\mathrm{2}} \left(\mathrm{2}{t}\right)−\mathrm{1}\right){dt}\:\:=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{16}}\:\int\left(\frac{\mathrm{1}+{ch}\left(\mathrm{4}{t}\right)}{\mathrm{2}}−\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}\int\:\left({ch}\left(\mathrm{4}{t}\right)−\mathrm{1}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}×\mathrm{4}}{sh}\left(\mathrm{4}{t}\right)−\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}{t}\:+{C} \\ $$$${sh}\left(\mathrm{4}{t}\right)\:=\frac{{e}^{\mathrm{4}{t}} −{e}^{−\mathrm{4}{t}} }{\mathrm{2}}\:\:{and}\:{t}={argsh}\left(\frac{{x}}{{a}}\right)\:={ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:\Rightarrow \\ $$$${sh}\left(\mathrm{4}{t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}\left\{\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right\}\:\Rightarrow \\ $$$${I}\:=\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{8}×\mathrm{32}}\left\{\:\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{\mathrm{4}} −\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)^{−\mathrm{4}} \right\} \\ $$$$−\frac{{a}^{\mathrm{3}} \mid{a}\mid}{\mathrm{32}}{ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:+{C}\:. \\ $$

Commented by Power last updated on 20/Feb/20

sir x=a tanu solution

$$\mathrm{sir}\:\:\mathrm{x}=\mathrm{a}\:\mathrm{tanu}\:\:\:\mathrm{solution} \\ $$

Commented by john santu last updated on 20/Feb/20

∫ x (x(√(a^2 +x^2 ))) dx = I u = x ⇒du = dx v = ∫ x(√(a^2 +x^2 )) dx = (1/2)∫ (√(a^2 +x^2 )) d(a^2 +x^2 ) = (1/3)(a^2 +x^2 )^(3/2) I = (1/3)x(a^2 +x^2 )^(3/2) −∫ (a^2 +x^2 )^(3/2) dx J = ∫ (a^2 +x^2 )^(3/2) dx let x = a tan θ ⇒dx = asec θ dθ J = ∫a^2 sec^4 θ dθ J = a^2 ∫ (1+tan^2 θ) d(tan θ) J = a^2 (tan θ+(1/3)tan^3 θ)+c J = a^2 ((x/a)+(x^3 /(3a^3 )))+c = ax + (1/3)(x^3 /a)+c ∴ I −J = (1/3)(√((a^2 +x^2 )^3 )) − ax−(x^3 /(3a)) + c

$$\int\:{x}\:\left({x}\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\right)\:{dx}\:=\:{I} \\ $$$${u}\:=\:{x}\:\Rightarrow{du}\:=\:{dx} \\ $$$${v}\:=\:\int\:{x}\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }\:{d}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$${I}\:=\:\frac{\mathrm{1}}{\mathrm{3}}{x}\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} −\int\:\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dx} \\ $$$${J}\:=\:\int\:\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:{dx} \\ $$$${let}\:{x}\:=\:{a}\:\mathrm{tan}\:\theta\:\Rightarrow{dx}\:=\:{a}\mathrm{sec}\:\theta\:{d}\theta \\ $$$${J}\:=\:\int{a}^{\mathrm{2}} \mathrm{sec}\:^{\mathrm{4}} \theta\:{d}\theta\: \\ $$$${J}\:=\:{a}^{\mathrm{2}} \int\:\left(\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} \theta\right)\:{d}\left(\mathrm{tan}\:\theta\right) \\ $$$${J}\:=\:{a}^{\mathrm{2}} \left(\mathrm{tan}\:\theta+\frac{\mathrm{1}}{\mathrm{3}}\mathrm{tan}\:^{\mathrm{3}} \theta\right)+{c} \\ $$$${J}\:=\:{a}^{\mathrm{2}} \left(\frac{{x}}{{a}}+\frac{{x}^{\mathrm{3}} }{\mathrm{3}{a}^{\mathrm{3}} }\right)+{c}\:=\:{ax}\:+\:\frac{\mathrm{1}}{\mathrm{3}}\frac{{x}^{\mathrm{3}} }{{a}}+{c} \\ $$$$\therefore\:{I}\:−{J}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\sqrt{\left({a}^{\mathrm{2}} +{x}^{\mathrm{2}} \right)^{\mathrm{3}} }\:−\:{ax}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}{a}}\:+\:{c} \\ $$

Commented by Power last updated on 20/Feb/20

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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