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Question Number 28702 by students last updated on 29/Jan/18

solve integration  (1/(√((x−α)(β−x))))  .

$${solve}\:{integration}\:\:\frac{\mathrm{1}}{\sqrt{\left({x}−\alpha\right)\left(\beta−{x}\right)}}\:\:. \\ $$

Commented by abdo imad last updated on 29/Jan/18

let use the ch. x= ((α−β)/2)t +((α+β)/2)so  x−α= ((α−β)/2)t +((α+β −2α)/2)=((α−β)/2) (t−1) and   β−x=β−((α+β)/2) −((α−β)/2)t =((β−α)/2) −((α−β)/2)t  = ((α−β)/2)(−1−t) so   I=∫  ? (dx/(√((x−α)(β−x))))= ∫      (1/(√(((α−β)/2)(t−1)((α−β)/2)(−1−1−t))))((α−β)/2)dt  =(((α−β)/2)/(∣((α−β)/2)∣)) ∫           (dt/(√(1−t^2 )))= ξ arcsin(t)+k  = ξ arcsin ( (2/(α−β))(x− ((α+β)/2)))  if α≠β and ξ^2 =1  and we mudt study the case α=β....

$${let}\:{use}\:{the}\:{ch}.\:{x}=\:\frac{\alpha−\beta}{\mathrm{2}}{t}\:+\frac{\alpha+\beta}{\mathrm{2}}{so} \\ $$$${x}−\alpha=\:\frac{\alpha−\beta}{\mathrm{2}}{t}\:+\frac{\alpha+\beta\:−\mathrm{2}\alpha}{\mathrm{2}}=\frac{\alpha−\beta}{\mathrm{2}}\:\left({t}−\mathrm{1}\right)\:{and}\: \\ $$$$\beta−{x}=\beta−\frac{\alpha+\beta}{\mathrm{2}}\:−\frac{\alpha−\beta}{\mathrm{2}}{t}\:=\frac{\beta−\alpha}{\mathrm{2}}\:−\frac{\alpha−\beta}{\mathrm{2}}{t} \\ $$$$=\:\frac{\alpha−\beta}{\mathrm{2}}\left(−\mathrm{1}−{t}\right)\:{so}\: \\ $$$${I}=\int\:\:?\:\frac{{dx}}{\sqrt{\left({x}−\alpha\right)\left(\beta−{x}\right)}}=\:\int\:\:\:\:\:\:\frac{\mathrm{1}}{\sqrt{\frac{\alpha−\beta}{\mathrm{2}}\left({t}−\mathrm{1}\right)\frac{\alpha−\beta}{\mathrm{2}}\left(−\mathrm{1}−\mathrm{1}−{t}\right)}}\frac{\alpha−\beta}{\mathrm{2}}{dt} \\ $$$$=\frac{\frac{\alpha−\beta}{\mathrm{2}}}{\mid\frac{\alpha−\beta}{\mathrm{2}}\mid}\:\int\:\:\:\:\:\:\:\:\:\:\:\frac{{dt}}{\sqrt{\mathrm{1}−{t}^{\mathrm{2}} }}=\:\xi\:{arcsin}\left({t}\right)+{k} \\ $$$$=\:\xi\:{arcsin}\:\left(\:\frac{\mathrm{2}}{\alpha−\beta}\left({x}−\:\frac{\alpha+\beta}{\mathrm{2}}\right)\right)\:\:{if}\:\alpha\neq\beta\:{and}\:\xi^{\mathrm{2}} =\mathrm{1} \\ $$$${and}\:{we}\:{mudt}\:{study}\:{the}\:{case}\:\alpha=\beta.... \\ $$

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