find-0-1-ln-x-x-1-x-3-2-dx-

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

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)

Facebook Tweet Pin