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x-2-4-1-2-dx-trig-substitution-only-

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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
∫(x2−4)1/2dxtrigsubstitutiononly
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 .
letI=∫x2−4dxweusethechangementx=2ch(t)⇒I=∫4ch2t−4(2sh(t)dt=4∫ch2t−1sh(t)dt=4∫sh2tdt=4∫ch(2t)−12dt=2∫ch(2t)dt−2t=sh(2t)−2t=2sh(t)ch(t)−2t=xch2t−1−2t=xx24−1−2t=x2x2−4−2argch(x2)=x2x2−4−2ln(x2+x24−1)=x2x2−4−2ln(x+x2−42)+c0=x2x2−4−2ln(x+x2−4)+4ln(2)+c0⇒∫x2−4dx=x2x2−4−2ln(x+x2−4)+C.

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