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Question Number 62251 by hovea cw last updated on 18/Jun/19

∫(x^2 −4)^(1/2) dx trig substitution only

$$\int\left(\mathrm{x}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{1}/\mathrm{2}} \mathrm{dx} \\ $$$$\mathrm{trig}\:\mathrm{substitution}\:\mathrm{only} \\ $$

Commented by maxmathsup by imad last updated on 18/Jun/19

let I =∫ (√(x^2 −4))dx we use the changement x =2ch(t) ⇒ I =∫(√(4ch^2 t−4())2sh(t)dt =4 ∫ (√(ch^2 t−1))sh(t)dt =4 ∫ sh^2 tdt =4 ∫ ((ch(2t)−1)/2)dt =2 ∫ ch(2t)dt−2t = sh(2t)−2t =2sh(t)ch(t)−2t =x(√(ch^2 t−1))−2t =x(√((x^2 /4)−1))−2t =(x/2)(√(x^2 −4))−2argch((x/2)) =(x/2)(√(x^2 −4)) −2 ln((x/2) +(√((x^2 /4)−1))) =(x/2)(√(x^2 −4)) −2ln(((x+(√(x^2 −4)))/2))+c_0 =(x/2)(√(x^2 −4)) −2ln(x+(√(x^2 −4))) +4ln(2) +c_0 ⇒ ∫ (√(x^2 −4))dx =(x/2)(√(x^2 −4)) −2ln(x+(√(x^2 −4))) +C .

$${let}\:{I}\:=\int\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}{dx}\:\:\:{we}\:{use}\:{the}\:{changement}\:{x}\:=\mathrm{2}{ch}\left({t}\right)\:\Rightarrow \\ $$$${I}\:=\int\sqrt{\mathrm{4}{ch}^{\mathrm{2}} {t}−\mathrm{4}\left(\right.}\mathrm{2}{sh}\left({t}\right){dt}\:=\mathrm{4}\:\int\:\:\sqrt{{ch}^{\mathrm{2}} {t}−\mathrm{1}}{sh}\left({t}\right){dt}\:=\mathrm{4}\:\int\:{sh}^{\mathrm{2}} {tdt} \\ $$$$=\mathrm{4}\:\int\:\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}{dt}\:=\mathrm{2}\:\int\:{ch}\left(\mathrm{2}{t}\right){dt}−\mathrm{2}{t}\:\:=\:{sh}\left(\mathrm{2}{t}\right)−\mathrm{2}{t} \\ $$$$=\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)−\mathrm{2}{t}\:={x}\sqrt{{ch}^{\mathrm{2}} {t}−\mathrm{1}}−\mathrm{2}{t}\:={x}\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}−\mathrm{2}{t}\:=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}−\mathrm{2}{argch}\left(\frac{{x}}{\mathrm{2}}\right) \\ $$$$=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:−\mathrm{2}\:{ln}\left(\frac{{x}}{\mathrm{2}}\:+\sqrt{\frac{{x}^{\mathrm{2}} }{\mathrm{4}}−\mathrm{1}}\right)\:=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:−\mathrm{2}{ln}\left(\frac{{x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}}{\mathrm{2}}\right)+{c}_{\mathrm{0}} \\ $$$$=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:\:−\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)\:+\mathrm{4}{ln}\left(\mathrm{2}\right)\:+{c}_{\mathrm{0}} \:\Rightarrow \\ $$$$\int\:\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}{dx}\:=\frac{{x}}{\mathrm{2}}\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\:−\mathrm{2}{ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{4}}\right)\:+{C}\:. \\ $$

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