((⊚BeMath⊚)/) ∫ x (√(x/(2a−x))) dx ?


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Question Number 107965 by bemath last updated on 13/Aug/20

$\frac{⊚ B e M a t h ⊚}{}$ $\int x \sqrt{\frac{x}{2 a - x}} d x ?$
	Answered by bobhans last updated on 13/Aug/20

	$\frac{⋱ BOBHANS \iddots}{Π}$ $I = \int x \sqrt{\frac{x}{2 a - x}} dx [set x = a - acos 2 t]$ $I = \int a (1 - \cos 2 t) \sqrt{\frac{a (1 - \cos 2 t)}{a (1 + \cos 2 t)}} (2 asin 2 t) dt$ $I = {2 a}^{2} \int (1 - \cos 2 t) \sin 2 t \sqrt{\frac{2 \sin^{2} t}{2 \cos^{2} t}} dt$ $I = {2 a}^{2} \int (1 - \cos 2 t) . 2 \sin t \cos t . \frac{\sin t}{\cos t} dt$ $I = {4 a}^{2} \int (1 - \cos 2 t) . \sin^{2} t dt$ $I = {4 a}^{2} \int (1 - \cos 2 t) (\frac{1 - \cos 2 t}{2}) dt$ $I = {2 a}^{2} \int (1 - 2 \cos 2 t + \cos^{2} 2 t) dt$ $I = {2 a}^{2} [t - \sin 2 t + \int (\frac{1 + \cos 4 t}{2}) dt]$ $I = {2 a}^{2} t - {2 a}^{2} \sin 2 t + a^{2} t + \frac{1}{4} a^{2} \sin 4 t + C$ $I = \frac{3}{2} a^{2} \cos^{- 1} (\frac{a - x}{a}) + \frac{1}{4} a^{2} \sin 4 t - {2 a}^{2} \sin 2 t + C$ $where \cos 2 t = \frac{a - x}{a}$
	Answered by john santu last updated on 13/Aug/20

	$\frac{♣ JS ♠}{⧫}$ $s e t x = 2 a \sin^{2} υ \Rightarrow d x = 2 a \sin 2 v d υ$ $I = \int 2 a \sin^{2} υ \sqrt{\frac{2 a \sin^{2} υ}{2 a - 2 a \sin^{2} υ}}$ $(4 a \sin υ \cos υ) d v$ $I = \int 2 a \sin^{2} υ . \frac{\sin υ}{\cos υ} . (4 a \sin υ \cos υ) d υ$ $I = 8 a^{2} \int \sin^{4} υ d υ$ $I = 8 a^{2} \int {(\frac{1 - \cos 2 υ}{2})}^{2} d υ$ $I = 2 a^{2} \int (1 - 2 \cos 2 υ + \frac{1 + \cos 4 υ}{2}) d υ$ $I = 2 a^{2} \int (\frac{3}{2} - 2 \cos 2 υ + \cos 4 υ) d υ$ $I = 2 a^{2} (\frac{3 υ}{2} - \sin 2 υ + \frac{1}{4} \sin 4 υ) + C$ $I = 3 a^{2} υ - 2 a^{2} \sin 2 υ + \frac{a^{2}}{2} \sin 4 υ + C$
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