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Question Number 169559 by TOTTI last updated on 03/May/22

Commented by cortano1 last updated on 03/May/22

$2^{n d} w a y$ $\begin{array}{ccc} x & {(1 - x)}^{\frac{1}{2}} \\ 1 & - \frac{2}{3} {(1 - x)}^{\frac{3}{2}} \\ 0 & - \frac{4}{15} {(1 - x)}^{\frac{5}{2}} \end{array}$ $I = \int x \sqrt{1 - x} d x = - \frac{2}{3} x \sqrt{{(1 - x)}^{3}} + \frac{4}{15} \sqrt{{(1 - x)}^{5}} + c$
	Answered by greougoury555 last updated on 03/May/22
	$∫ x(√(1−x)) dx let (√(1−x)) = u → { ((x=1−u^2 )),((dx=−2u du)) :} I=∫ (1−u^2 )(u)(−2u)du =−2∫(1−u^2 )u^2 du =−2∫(u^2 −u^4 )du =−2[ (1/3)u^3 −(1/5)u^5 ]+c =−(2/(15))u^3 (5−3u^2 )+c =−(2/(15)) (√((1−x)^3 )) (5−3(1−x))+c =−(2/(15)) (3x+2)(√((1−x)^3 )) + c$
	$\int x \sqrt{1 - x} d x$ $l e t \sqrt{1 - x} = u \to {\begin{cases} x = 1 - u^{2} \\ d x = - 2 u d u \end{cases}$ $I = \int (1 - u^{2}) (u) (- 2 u) d u$ $= - 2 \int (1 - u^{2}) u^{2} d u$ $= - 2 \int (u^{2} - u^{4}) d u$ $= - 2 [\frac{1}{3} u^{3} - \frac{1}{5} u^{5}] + c$ $= - \frac{2}{15} u^{3} (5 - 3 u^{2}) + c$ $= - \frac{2}{15} \sqrt{{(1 - x)}^{3}} (5 - 3 (1 - x)) + c$ $= - \frac{2}{15} (3 x + 2) \sqrt{{(1 - x)}^{3}} + c$
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