calculate ∫ (dx/(√((x−1)(2−x))))


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

$c a l c u l a t e \int \frac{d x}{\sqrt{(x - 1) (2 - x)}}$
Commented by Prithwish sen last updated on 08/Jul/19

$put x - l = z^{2} \Rightarrow dx = 2 zdz$ $\int \frac{2 zdz}{z \sqrt{3 - z^{2}}} = 2 \int \frac{dz}{\sqrt{(\sqrt{{3)}^{2}} - z^{2}}}$
Commented by mathmax by abdo last updated on 08/Jul/19
$let I =∫ (dx/(√((x−1)(2−x)))) we have (x−1)(2−x)=2x−x^2 −2 +x −x^2 +3x−2 =−(x^2 −3x+2) =−(x^2 −2(3/2)x +(9/4) +2−(9/4)) =−{(x−(3/2))^2 −(1/4)} =(1/4)−(x−(3/2))^2 for that we use the changement x−(3/2) =((sint)/2) ⇒I = ∫ ((costdt)/(2(1/2)(√(1−sin^2 t)))) = ∫ ((cost)/(cost))dt +c =∫dt+c = t+c =arcsin(2x−3) +c .$
$l e t I = \int \frac{d x}{\sqrt{(x - 1) (2 - x)}} w e h a v e (x - 1) (2 - x) = 2 x - x^{2} - 2 + x$ $- x^{2} + 3 x - 2 = - (x^{2} - 3 x + 2) = - (x^{2} - 2 \frac{3}{2} x + \frac{9}{4} + 2 - \frac{9}{4})$ $= - {{(x - \frac{3}{2})}^{2} - \frac{1}{4}} = \frac{1}{4} - {(x - \frac{3}{2})}^{2} f o r t h a t w e u s e t h e c h a n g e m e n t$ $x - \frac{3}{2} = \frac{s i n t}{2} \Rightarrow I = \int \frac{c o s t d t}{2 \frac{1}{2} \sqrt{1 - {s i n}^{2} t}} = \int \frac{c o s t}{c o s t} d t + c$ $= \int d t + c = t + c = a r c s i n (2 x - 3) + c .$
	Answered by MJS last updated on 08/Jul/19

	$\int \frac{d x}{\sqrt{(x - 1) (2 - x)}} =$ $[t = \arccos (2 x - 3) \to d x = - \sqrt{(x - 1) (2 - x)} d t]$ $= - \int d t = - t = - \arccos (2 x - 3) + C$
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