find ∫ x(√((1−(√x))/(1+(√x))))dx


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Question Number 46740 by math khazana by abdo last updated on 30/Oct/18

$f i n d \int x \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x$
Commented by behi83417@gmail.com last updated on 31/Oct/18

$= \int x . \sqrt{\frac{(1 - \sqrt{x}) (1 - \sqrt{x})}{(1 + \sqrt{x}) (1 - \sqrt{x})}} d x =$ $= \int x . \frac{1 - \sqrt{x}}{\sqrt{1 - x}} d x = \int [\frac{x}{\sqrt{1 - x}} - \frac{x \sqrt{x}}{\sqrt{1 - x}}] d x =$ $[x = {c o s}^{2} t \Rightarrow d x = - 2 c o s t . s i n t d t]$ $= \int [\frac{{c o s}^{2} t}{s i n t} - \frac{{c o s}^{3} t}{s i n t}] (- 2 c o s t . s i n t) d t =$ $= 2 \int ({c o s}^{4} t - {c o s}^{3} t) d t =$ $= 2 \int [{(\frac{1 + c o s 2 t}{2})}^{2} - (\frac{1}{4} c o s 3 t + \frac{3}{4} c o s t)] d t =$ $= \frac{1}{4} \int [3 + 4 c o s 2 t + c o s 4 t - 2 c o s 3 t + 6 c o s t] d t =$ $= \frac{3}{4} t + \frac{1}{2} s i n 2 t + \frac{1}{16} s i n 4 t - \frac{1}{6} s i n 3 t + \frac{3}{2} s i n t + C =$ $= \frac{3}{4} {c o s}^{- 1} \sqrt{x} + \sqrt{x} . \sqrt{1 - x} + \frac{1}{4} \sqrt{x} . \sqrt{1 - x} (2 x - 1)$ $- \frac{1}{6} [3 \sqrt{1 - x} - 4 (1 - x) \sqrt{1 - x}] + \frac{3}{2} \sqrt{1 - x} + C .$ $[s i n 4 t = 2 \sin 2 tcos 2 t = 4 sintcost . ({2 \cos}^{2} t - 1) =$ $= 4 \sqrt{x} . \sqrt{1 - x} . (2 x - 1)$ $\sin 3 t = 3 sint - {4 \sin}^{3} t = 3 \sqrt{1 - x} - 4 (1 - x) \sqrt{1 - x}$
Commented by maxmathsup by imad last updated on 31/Oct/18
$we have x(√((1−(√x))/(1+(√x))))=x(√(((1−(√x))^2 )/(1−x)))=x((1−(√x))/(√(1−x))) =(x/(√(1−x))) −((x(√x))/(√(1−x))) ⇒ I = ∫ ((xdx)/(√(1−x))) −∫ ((x(√x))/(√(1−x))) changement x=sin^2 θ give ∫ (x/(√(1−x)))dx = ∫ ((sin^2 θ)/(cosθ)) 2sinθ cosθ = 2 ∫ sin^3 θ dθ but sin^3 θ =(((e^(iθ ) −e^(−iθ) )/(2i)))^3 =−(1/(8i)) Σ_(k=0) ^3 C_3 ^k e^(ikθ) e^(−(3−k)iθ) =(i/8)Σ_(k=0) ^3 C_3 ^k (−1)^(3−k) e^(i(2k−3)θ) =(1/8){− e^(−i3θ) +3 e^(−iθ) −3 e^(iθ) + e^(i3θ) } =(i/8){2isin(3θ) −6i sin(θ)} =−(1/4)sin(3θ) +(3/4)sin(θ) ⇒ ∫ (x/(√(1−x)))dx =−(1/2) ∫ sin(3θ)dθ +(3/2) ∫ sinθ dθ +c_1 =(1/6) cos(3θ)−(3/2)cosθ +c_1 also the same changeent x=sin^2 θ give ∫ ((x(√x))/(√(1−x))) dx = ∫ ((sin^2 θ sinθ)/(cosθ)) 2sinθ cosθ dθ = 2 ∫ sin^4 θ dθ =2 ∫ (((1−cos(2θ))/2))^2 dθ =(1/2) ∫ (1−2cos(2θ) +((1+cos(4θ))/2))dθ =(θ/2) −(1/2)sin(2θ) +(θ/4) +(1/8)sin(4θ) +c_2 =((3θ)/4) −((sin(2θ))/2) +((sin(4θ))/8) +c_2 ⇒ I =(1/6)cos(3θ)−(3/2) cosθ +((3θ)/4) −((sin(2θ))/2) +((sin(4θ))/8) +C =(1/6) cos(3 arcsin((√x)))−(3/2) cos(arcsin((√x)))+(3/4) arcsin((√x))−(1/2)sin(2arcsin(√x)) +(1/8) sin(4 arcsin(√x)) +C .$
$w e h a v e x \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} = x \sqrt{\frac{{(1 - \sqrt{x})}^{2}}{1 - x}} = x \frac{1 - \sqrt{x}}{\sqrt{1 - x}} = \frac{x}{\sqrt{1 - x}} - \frac{x \sqrt{x}}{\sqrt{1 - x}} \Rightarrow$ $I = \int \frac{x d x}{\sqrt{1 - x}} - \int \frac{x \sqrt{x}}{\sqrt{1 - x}} c h a n g e m e n t x = {s i n}^{2} θ g i v e$ $\int \frac{x}{\sqrt{1 - x}} d x = \int \frac{{s i n}^{2} θ}{c o s θ} 2 s i n θ c o s θ = 2 \int {s i n}^{3} θ d θ b u t$ ${s i n}^{3} θ = {(\frac{e^{i θ} - e^{- i θ}}{2 i})}^{3} = - \frac{1}{8 i} \sum_{k = 0}^{3} C_{3}^{k} e^{i k θ} e^{- (3 - k) i θ}$ $= \frac{i}{8} \sum_{k = 0}^{3} C_{3}^{k} {(- 1)}^{3 - k} e^{i (2 k - 3) θ} = \frac{1}{8} {- e^{- i 3 θ} + 3 e^{- i θ} - 3 e^{i θ} + e^{i 3 θ}}$ $= \frac{i}{8} {2 i s i n (3 θ) - 6 i s i n (θ)} = - \frac{1}{4} s i n (3 θ) + \frac{3}{4} s i n (θ) \Rightarrow$ $\int \frac{x}{\sqrt{1 - x}} d x = - \frac{1}{2} \int s i n (3 θ) d θ + \frac{3}{2} \int s i n θ d θ + c_{1}$ $= \frac{1}{6} c o s (3 θ) - \frac{3}{2} c o s θ + c_{1} a l s o t h e s a m e c h a n g e e n t x = {s i n}^{2} θ g i v e$ $\int \frac{x \sqrt{x}}{\sqrt{1 - x}} d x = \int \frac{{s i n}^{2} θ s i n θ}{c o s θ} 2 s i n θ c o s θ d θ = 2 \int {s i n}^{4} θ d θ$ $= 2 \int {(\frac{1 - c o s (2 θ)}{2})}^{2} d θ = \frac{1}{2} \int (1 - 2 c o s (2 θ) + \frac{1 + c o s (4 θ)}{2}) d θ$ $= \frac{θ}{2} - \frac{1}{2} s i n (2 θ) + \frac{θ}{4} + \frac{1}{8} s i n (4 θ) + c_{2}$ $= \frac{3 θ}{4} - \frac{s i n (2 θ)}{2} + \frac{s i n (4 θ)}{8} + c_{2} \Rightarrow$ $I = \frac{1}{6} c o s (3 θ) - \frac{3}{2} c o s θ + \frac{3 θ}{4} - \frac{s i n (2 θ)}{2} + \frac{s i n (4 θ)}{8} + C$ $= \frac{1}{6} c o s (3 a r c s i n (\sqrt{x})) - \frac{3}{2} c o s (a r c s i n (\sqrt{x})) + \frac{3}{4} a r c s i n (\sqrt{x}) - \frac{1}{2} s i n (2 a r c s i n \sqrt{x})$ $+ \frac{1}{8} s i n (4 a r c s i n \sqrt{x}) + C .$
	Answered by MJS last updated on 31/Oct/18

	$\int x \sqrt{\frac{1 - \sqrt{x}}{1 + \sqrt{x}}} d x =$ $[t = \sqrt{1 - \sqrt{x}} \to d x = - 4 \sqrt{x (1 - \sqrt{x})} d t]$ $= 4 \int \frac{t^{2} {(1 - t^{2})}^{3}}{\sqrt{2 - t^{2}}} d t =$ $[t = \sqrt{2} \sin u \to u = \arcsin \frac{t \sqrt{2}}{2}; d t = d u \sqrt{2} \cos u]$ $= 4 \int \frac{\sqrt{8} \cos u \sin^{2} u {({2 \sin}^{2} u - 1)}^{3}}{\sqrt{2 - {2 \sin}^{2} u}} d u =$ $= - 8 \int \sin^{2} u {(1 - {2 \sin}^{2} u)}^{3} d u =$ $= - 8 \int (- {8 \sin}^{8} u + {12 \sin}^{6} u - {6 \sin}^{4} u + \sin^{2} u) d u =$ $= \frac{1}{2} \int \cos 8 u d u - \int \cos 6 u d u + 2 \int \cos 4 u d u - 3 \int \cos 2 u d u + \frac{3}{2} \int d u =$ $= \frac{1}{16} \sin 8 u - \frac{1}{6} \sin 6 u + \frac{1}{2} \sin 4 u - \frac{3}{2} \sin 2 u + \frac{3}{2} u =$ $= . . . hard but possible, leads to . . . =$ $= \frac{3}{2} \arcsin \sqrt{\frac{1 - \sqrt{x}}{2}} + (\frac{1}{2} x^{\frac{3}{2}} - \frac{2}{3} x + \frac{3}{4} x^{\frac{1}{2}} - \frac{4}{3}) \sqrt{1 - x} + C$
	Commented by maxmathsup by imad last updated on 31/Oct/18

	$t h a n k y o u s i r .$
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