calculate-x-3-2-x-dx-

Question Number 63720 by mathmax by abdo last updated on 08/Jul/19

c a l c u l a t e \int \sqrt{(x - 3) (2 - x)} d x

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

\int \sqrt{- x^{2} + 5 x - 6} dx = \int \sqrt{{(\frac{1}{2})}^{2} - {(x - \frac{5}{2})}^{2}} dx

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

w e h a v e (x - 3) (2 - x) = 2 x - x^{2} - 6 + 3 x = - x^{2} + 5 x - 6

= - (x^{2} - 5 x + 6) = - {x^{2} - 2 \frac{5}{2} x + \frac{25}{4} + 6 - \frac{25}{4}}

= - {{(x - \frac{5}{2})}^{2} - \frac{1}{4}} = \frac{1}{4} - {(x - \frac{5}{2})}^{2} l e t u s e t h e c h a n g e m e n t

x - \frac{5}{2} = \frac{s i n t}{2} \Rightarrow

\int \sqrt{(x - 3) (2 - x)} d x = \int \frac{1}{2} \sqrt{1 - {s i n}^{2} t} \frac{c o s t}{2} d t

= \frac{1}{4} \int {c o s}^{2} t d t = \frac{1}{4} \int \frac{1 + c o s (2 t)}{2} d t = \frac{1}{8} \int (1 + c o s (2 t)) d t

= \frac{t}{8} + \frac{1}{16} s i n (2 t) + c = \frac{t}{8} + \frac{1}{8} s i n t c o s t + c

= a r c s i n (2 x - 5) + \frac{1}{8} (2 x - 5) \sqrt{1 - {(2 x - 5)}^{2}} + c .

Answered by MJS last updated on 08/Jul/19

\int \sqrt{(x - 3) (2 - x)} d x =

[t = \arccos (2 x - 5) \to d x = - \sqrt{(x - 3) (2 - x)} d t]

= - \frac{1}{4} \int \sin^{2} t d t = - \frac{1}{8} \int d t + \frac{1}{8} \int \cos 2 t d t =

= - \frac{1}{8} t + \frac{1}{16} \sin 2 t = - \frac{1}{8} t + \frac{1}{8} \sin t \cos t =

= - \frac{1}{8} \arccos (2 x - 5) + \frac{1}{4} (2 x - 5) \sqrt{(x - 3) (2 - x)} + C

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