find ∫_0 ^1 ((ln(x))/((√x)(1−x)^(3/2) ))dx


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Question Number 50422 by Abdo msup. last updated on 16/Dec/18

$f i n d \int_{0}^{1} \frac{l n (x)}{\sqrt{x} {(1 - x)}^{\frac{3}{2}}} d x$
Commented by Abdo msup. last updated on 23/Dec/18
$changement x =sin^2 θ give ∫_0 ^1 ((ln(x))/((√x)(1−x)^(3/2) ))dx =∫_0 ^(π/2) ((2ln(sinθ))/(sinθ(cos^2 θ)^(3/2) )) 2sinθ cosθ dθ =4∫_0 ^(π/2) ((ln(sinθ))/(cos^2 θ)) dθ by psrts u^′ =(1/(cos^2 θ)) and v=ln(sinθ) =4 { [tanθ ln(sinθ)]_0 ^(π/2) −∫_0 ^(π/2) tanθ ((cosθ)/(sinθ))dθ} =−4 ∫_0 ^(π/2) dθ =−2π .rest to prove that lim_(θ→0) tanθ ln(sinθ)=0 and lim_(θ→(π/2)) tanθ ln(sinθ)=0 for x∈v(0) tanθ∼ θ ,ln(sinθ)∼ln(θ) ⇒ tanθ ln(sinθ)∼θ lnθ →0 (θ→0) ch. θ =(π/2)−t give tanθ ln(sinθ) =tan((π/2)−t)ln(cost) =((ln(cost))/(tan(t))) but cost ∼ 1−(t^2 /2) and ln(cost)∼−(t^2 /2) tan(t)∼ t ⇒((ln(cost))/(tan(t))) ∼ −(t/2) →0(t→0) ⇒the result is proved.$
$c h a n g e m e n t x = {s i n}^{2} θ g i v e$ $\int_{0}^{1} \frac{l n (x)}{\sqrt{x} {(1 - x)}^{\frac{3}{2}}} d x = \int_{0}^{\frac{π}{2}} \frac{2 l n (s i n θ)}{s i n θ {({c o s}^{2} θ)}^{\frac{3}{2}}} 2 s i n θ c o s θ d θ$ $= 4 \int_{0}^{\frac{π}{2}} \frac{l n (s i n θ)}{{c o s}^{2} θ} d θ b y p s r t s u^{^{'}} = \frac{1}{{c o s}^{2} θ} a n d v = l n (s i n θ)$ $= 4 {{[t a n θ l n (s i n θ)]}_{0}^{\frac{π}{2}} - \int_{0}^{\frac{π}{2}} t a n θ \frac{c o s θ}{s i n θ} d θ}$ $= - 4 \int_{0}^{\frac{π}{2}} d θ = - 2 π . r e s t t o p r o v e t h a t$ ${l i m}_{θ \to 0} t a n θ l n (s i n θ) = 0 a n d {l i m}_{θ \to \frac{π}{2}} t a n θ l n (s i n θ) = 0$ $f o r x \in v (0) t a n θ \sim θ, l n (s i n θ) \sim l n (θ) \Rightarrow$ $t a n θ l n (s i n θ) \sim θ l n θ \to 0 (θ \to 0)$ $c h . θ = \frac{π}{2} - t g i v e t a n θ l n (s i n θ) = t a n (\frac{π}{2} - t) l n (c o s t)$ $= \frac{l n (c o s t)}{t a n (t)} b u t c o s t \sim 1 - \frac{t^{2}}{2} a n d l n (c o s t) \sim - \frac{t^{2}}{2}$ $t a n (t) \sim t \Rightarrow \frac{l n (c o s t)}{t a n (t)} \sim - \frac{t}{2} \to 0 (t \to 0) \Rightarrow t h e r e s u l t$ $i s p r o v e d .$
	Answered by Smail last updated on 27/Dec/18

	$b y p a r t s$ $u = l n x \Rightarrow u^{'} = \frac{1}{x}$ $v^{'} = \frac{1}{\sqrt{x} {(1 - x)}^{3 / 2}} \Rightarrow v = \frac{2 \sqrt{x}}{\sqrt{1 - x}}$ $I = \int_{0}^{1} \frac{l n x}{\sqrt{x} {(1 - x)}^{3 / 2}} d x = 2 {[\frac{\sqrt{x} l n x}{\sqrt{1 - x}}]}_{0}^{1} - 2 \int_{0}^{1} \frac{d x}{\sqrt{x} \sqrt{1 - x}}$ $l e t t^{2} = x \Rightarrow 2 t d t = d x$ $I = 0 - 4 \int_{0}^{1} \frac{t d t}{t \sqrt{1 - t^{2}}}$ $= - 4 {[{s i n}^{- 1} t]}_{0}^{1} = - 4 (\frac{π}{2} + 2 k π - k π)$ $= - 4 (\frac{π}{2} + k π)$
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