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Question Number 45231 by maxmathsup by imad last updated on 10/Oct/18

find ∫ (√((x−1)(3−x)))dx

$${find}\:\int\:\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{3}−{x}\right)}{dx} \\ $$

Commented by maxmathsup by imad last updated on 12/Oct/18

let A =∫ (√((x−1)(3−x)))dx ⇒A =∫(√(3x−x^2 −3+x))dx=∫(√(−x^2 +4x−3))dx =∫(√(−(x^2 −4x +4−4)−3))dx=∫(√(1−(x−2)^2 ))dxchangement x−2=sint give A =∫ costcost dt =∫ ((1+cos(2t))/2)dt =(t/2) +(1/4)sin(2t)+c =(t/2) +(1/2)sin(t)cos(t) +c =(1/2) arcsin(x−2) +((x−2)/2)(√(1−(x−2)^2 ))+c

$${let}\:{A}\:=\int\:\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{3}−{x}\right)}{dx}\:\Rightarrow{A}\:=\int\sqrt{\mathrm{3}{x}−{x}^{\mathrm{2}} −\mathrm{3}+{x}}{dx}=\int\sqrt{−{x}^{\mathrm{2}} \:+\mathrm{4}{x}−\mathrm{3}}{dx} \\ $$$$=\int\sqrt{−\left({x}^{\mathrm{2}} −\mathrm{4}{x}\:+\mathrm{4}−\mathrm{4}\right)−\mathrm{3}}{dx}=\int\sqrt{\mathrm{1}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }{dxchangement} \\ $$$${x}−\mathrm{2}={sint}\:{give}\:{A}\:=\int\:\:{costcost}\:{dt}\:=\int\:\frac{\mathrm{1}+{cos}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sin}\left(\mathrm{2}{t}\right)+{c}\:=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sin}\left({t}\right){cos}\left({t}\right)\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:{arcsin}\left({x}−\mathrm{2}\right)\:+\frac{{x}−\mathrm{2}}{\mathrm{2}}\sqrt{\mathrm{1}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }+{c} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 11/Oct/18

∫(√(3x−x^2 −3+x dx)) ∫(√(−3+4x−x^2 )) dx ∫(√(−3+4−4+4x−x^2 )) dx ∫(√(1−(x−2)^2 )) dx use formula ∫(√(a^2 −x^2 )) dx ((x−2)/2)(√(1−(x−2)^2 )) +(1/2)sin^(−1) (((x−2)/1))+c

$$\int\sqrt{\mathrm{3}{x}−{x}^{\mathrm{2}} −\mathrm{3}+{x}\:{dx}} \\ $$$$\int\sqrt{−\mathrm{3}+\mathrm{4}{x}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$\int\sqrt{−\mathrm{3}+\mathrm{4}−\mathrm{4}+\mathrm{4}{x}−{x}^{\mathrm{2}} }\:{dx} \\ $$$$\int\sqrt{\mathrm{1}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:{dx} \\ $$$${use}\:{formula}\:\int\sqrt{{a}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:{dx} \\ $$$$\frac{{x}−\mathrm{2}}{\mathrm{2}}\sqrt{\mathrm{1}−\left({x}−\mathrm{2}\right)^{\mathrm{2}} }\:+\frac{\mathrm{1}}{\mathrm{2}}{sin}^{−\mathrm{1}} \left(\frac{{x}−\mathrm{2}}{\mathrm{1}}\right)+{c} \\ $$

Answered by ajfour last updated on 11/Oct/18

let x−2 = t ∫(√((x−1)(3−x)))dx = ∫(√((t+1)(1−t)))dt = ∫(√(1−t^2 )) dt = (t/2)(√(1−t^2 ))+(1/2)sin^(−1) t +c = (((x−2)/2))(√((x−1)(3−x)))+(1/2)sin^(−1) (x−2)+c .

$${let}\:\:{x}−\mathrm{2}\:=\:{t} \\ $$$$\int\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{3}−{x}\right)}{dx}\:=\:\int\sqrt{\left({t}+\mathrm{1}\right)\left(\mathrm{1}−{t}\right)}{dt} \\ $$$$\:\:=\:\int\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }\:{dt} \\ $$$$\:\:=\:\frac{{t}}{\mathrm{2}}\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} {t}\:+{c} \\ $$$$\:=\:\left(\frac{{x}−\mathrm{2}}{\mathrm{2}}\right)\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{3}−{x}\right)}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}^{−\mathrm{1}} \left({x}−\mathrm{2}\right)+{c}\:. \\ $$

Commented by Necxx last updated on 12/Oct/18

hmmm....I caught it.Thanks

$${hmmm}....{I}\:{caught}\:{it}.{Thanks} \\ $$

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