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Question Number 63721 by mathmax by abdo last updated on 08/Jul/19

calculate ∫ (dx/(√((x−1)(2−x))))

$${calculate}\:\int\:\:\frac{{dx}}{\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}} \\ $$

Commented by Prithwish sen last updated on 08/Jul/19

put x−l = z^2 ⇒dx = 2zdz ∫((2zdz)/(z(√(3−z^2 )))) = 2∫(dz/(√(((√(3)^2 )) −z^2 )))

$$\mathrm{put}\:\mathrm{x}−\mathrm{l}\:=\:\mathrm{z}^{\mathrm{2}} \:\:\:\Rightarrow\mathrm{dx}\:=\:\mathrm{2zdz} \\ $$$$\int\frac{\mathrm{2zdz}}{\mathrm{z}\sqrt{\mathrm{3}−\mathrm{z}^{\mathrm{2}} \:}}\:\:=\:\mathrm{2}\int\frac{\mathrm{dz}}{\sqrt{\left(\sqrt{\left.\mathrm{3}\right)^{\mathrm{2}} }\:−\mathrm{z}^{\mathrm{2}} \right.}}\:\:\:\: \\ $$

Commented by mathmax by abdo last updated on 08/Jul/19

let I =∫ (dx/(√((x−1)(2−x)))) we have (x−1)(2−x)=2x−x^2 −2 +x −x^2 +3x−2 =−(x^2 −3x+2) =−(x^2 −2(3/2)x +(9/4) +2−(9/4)) =−{(x−(3/2))^2 −(1/4)} =(1/4)−(x−(3/2))^2 for that we use the changement x−(3/2) =((sint)/2) ⇒I = ∫ ((costdt)/(2(1/2)(√(1−sin^2 t)))) = ∫ ((cost)/(cost))dt +c =∫dt+c = t+c =arcsin(2x−3) +c .

$${let}\:{I}\:=\int\:\:\frac{{dx}}{\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}}\:\:{we}\:{have}\:\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)=\mathrm{2}{x}−{x}^{\mathrm{2}} −\mathrm{2}\:+{x} \\ $$$$−{x}^{\mathrm{2}} \:+\mathrm{3}{x}−\mathrm{2}\:=−\left({x}^{\mathrm{2}} −\mathrm{3}{x}+\mathrm{2}\right)\:=−\left({x}^{\mathrm{2}} −\mathrm{2}\frac{\mathrm{3}}{\mathrm{2}}{x}\:+\frac{\mathrm{9}}{\mathrm{4}}\:+\mathrm{2}−\frac{\mathrm{9}}{\mathrm{4}}\right) \\ $$$$=−\left\{\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}\right\}\:=\frac{\mathrm{1}}{\mathrm{4}}−\left({x}−\frac{\mathrm{3}}{\mathrm{2}}\right)^{\mathrm{2}} \:{for}\:{that}\:{we}\:{use}\:{the}\:{changement} \\ $$$${x}−\frac{\mathrm{3}}{\mathrm{2}}\:=\frac{{sint}}{\mathrm{2}}\:\Rightarrow{I}\:=\:\int\:\:\:\frac{{costdt}}{\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\mathrm{1}−{sin}^{\mathrm{2}} {t}}}\:=\:\int\:\:\:\frac{{cost}}{{cost}}{dt}\:+{c} \\ $$$$=\int{dt}+{c}\:=\:{t}+{c}\:={arcsin}\left(\mathrm{2}{x}−\mathrm{3}\right)\:+{c}\:. \\ $$

Answered by MJS last updated on 08/Jul/19

∫(dx/(√((x−1)(2−x))))= [t=arccos (2x−3) → dx=−(√((x−1)(2−x)))dt] =−∫dt=−t=−arccos (2x−3) +C

$$\int\frac{{dx}}{\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}}= \\ $$$$\:\:\:\:\:\left[{t}=\mathrm{arccos}\:\left(\mathrm{2}{x}−\mathrm{3}\right)\:\rightarrow\:{dx}=−\sqrt{\left({x}−\mathrm{1}\right)\left(\mathrm{2}−{x}\right)}{dt}\right] \\ $$$$=−\int{dt}=−{t}=−\mathrm{arccos}\:\left(\mathrm{2}{x}−\mathrm{3}\right)\:+{C} \\ $$

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