Question-69597

Question Number 69597 by ozodbek last updated on 25/Sep/19

Commented by mathmax by abdo last updated on 25/Sep/19

generally let find f(a) =∫(√(x^2 +a^2 ))dx with a>0 changement x =ash(t) give f(a)=∫ach(t)ach(t)dt =a^2 ∫ ch^2 (t)dt =a^2 ∫((1+ch(2t))/2)dt =(a^2 /2)t +(a^2 /2) ∫ch(2t)dt =(a^2 /2)t + (a^2 /4)sh(2t)+c =(a^2 /2)t +(a^2 /2)sh(t)ch(t) +c we have t =argsh((x/a)) =ln((x/a)+(√(1+(x^2 /a^2 )))) =ln(((x+(√(x^2 +a^2 )))/a)) ⇒ f(a) =(a^2 /2)ln(((x+(√(x^2 +a^2 )))/a))+(a^2 /2)(x/a)(√(1+(x^2 /a^2 ))) +c =(a^2 /2)ln(x+(√(x^2 +a^2 )))−(a^2 /2)ln(a) +(x/2)(√(x^2 +a^2 )) +c =(a^2 /2)ln(x+(√(x^2 +a^2 ))) +(x/2)(√(x^2 +a^2 )) +C a=2 ⇒ ∫ (√(4+x^2 ))dx =2ln(x+(√(x^2 +4)))+(x/2)(√(x^2 +4)) +C

$${generally}\:{let}\:{find}\:{f}\left({a}\right)\:=\int\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }{dx}\:\:{with}\:{a}>\mathrm{0} \\ $$$${changement}\:{x}\:={ash}\left({t}\right)\:{give}\:{f}\left({a}\right)=\int{ach}\left({t}\right){ach}\left({t}\right){dt} \\ $$$$={a}^{\mathrm{2}} \int\:{ch}^{\mathrm{2}} \left({t}\right){dt}\:={a}^{\mathrm{2}} \int\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\:\int{ch}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:\:+\:\frac{{a}^{\mathrm{2}} }{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)+{c}\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{t}\:+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:+{c} \\ $$$${we}\:{have}\:{t}\:={argsh}\left(\frac{{x}}{{a}}\right)\:={ln}\left(\frac{{x}}{{a}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\right)\:={ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}}\right)\:\Rightarrow \\ $$$${f}\left({a}\right)\:=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}{{a}}\right)+\frac{{a}^{\mathrm{2}} }{\mathrm{2}}\frac{{x}}{{a}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:+{c} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)−\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({a}\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+{c} \\ $$$$=\frac{{a}^{\mathrm{2}} }{\mathrm{2}}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }\:+{C} \\ $$$${a}=\mathrm{2}\:\Rightarrow\:\int\:\sqrt{\mathrm{4}+{x}^{\mathrm{2}} }{dx}\:=\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)+\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\:+{C} \\ $$

Answered by MJS last updated on 25/Sep/19

∫(√(4+x^2 ))dx= [t=sinh^(−1) (x/2) → dx=(√(x^2 +4))dt] =4∫cosh^2 t dt=2∫dt+2∫cosh 2t dt= =2t+sinh 2t =2sinh^(−1) (x/2) +((x(√(x^2 +4)))/2) +C different path leads to 2ln (x+(√(x^2 +4))) +((x(√(x^2 +4)))/2) +C

$$\int\sqrt{\mathrm{4}+{x}^{\mathrm{2}} }{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{sinh}^{−\mathrm{1}} \:\frac{{x}}{\mathrm{2}}\:\rightarrow\:{dx}=\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}{dt}\right] \\ $$$$=\mathrm{4}\int\mathrm{cosh}^{\mathrm{2}} \:{t}\:{dt}=\mathrm{2}\int{dt}+\mathrm{2}\int\mathrm{cosh}\:\mathrm{2}{t}\:{dt}= \\ $$$$=\mathrm{2}{t}+\mathrm{sinh}\:\mathrm{2}{t}\:=\mathrm{2sinh}^{−\mathrm{1}} \:\frac{{x}}{\mathrm{2}}\:+\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:+{C} \\ $$$$\mathrm{different}\:\mathrm{path}\:\mathrm{leads}\:\mathrm{to} \\ $$$$\mathrm{2ln}\:\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}\right)\:+\frac{{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{4}}}{\mathrm{2}}\:+{C} \\ $$

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