∫_0 ^1 ((ln(x+(√(1−x^2 ))))/x)dx=(π^2 /(16))


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Question Number 136765 by Ñï= last updated on 25/Mar/21

$\int_{0}^{1} \frac{l n (x + \sqrt{1 - x^{2}})}{x} d x = \frac{π^{2}}{16}$
	Answered by snipers237 last updated on 25/Mar/21

	$l e t n a m e d i t A$ $b y s t a t i n g x = s i n t$ $A = \int_{0}^{\frac{π}{2}} \frac{l n (s i n t + c o s t)}{t a n t} d t = \int_{0}^{\frac{π}{2}} t a n t . l n (s i n t + c o s t) d t d u e t o \frac{π}{2} - t s t a b i l i t y$ $2 A = \int_{0}^{\frac{π}{2}} (t a n t + \frac{1}{t a n t}) l n (s i n t + c o s t) d t$ $2 A = \int_{0}^{\frac{π}{2}} \frac{2 l n (s i n t + c o s t)}{s i n 2 t} d t = \int_{0}^{\frac{π}{2}} \frac{l n (1 + s i n 2 t)}{s i n (2 t)} d t$ $4 A = \int_{0}^{π} \frac{l n (1 + s i n t)}{s i n t} d t$ $L e t g (a) = \int_{0}^{π} \frac{l n (1 + a s i n t)}{s i n t} d t$ $g^{^{'}} (a) = \int_{0}^{π} \frac{s i n t d t}{s i n t (1 + a s i n t)} = \int_{0}^{π} \frac{a^{- 1}}{a^{- 1} + s i n t} d t$ $g^{'} (a) = \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{a^{- 1} d u}{a^{- 1} + c o s u} = 2 \int_{0}^{\frac{π}{2}} \frac{a^{- 1}}{a^{- 1} + c o s u} d u = 2 a^{- 1} . \frac{2 π}{\sqrt{a^{- 2} - 1}} w i t h u = \frac{π}{2} - t$ $g^{'} (a) = \frac{4 π}{\sqrt{1 - a^{2}}} . S o g (a) = 4 π a r c s i n (a) c a u s e g (0) = 0$ $T h e n g (1) = 4 π a r c s i n (1) = 2 π^{2}$ $a n d A = \frac{π^{2}}{2}$
	Answered by Ñï= last updated on 26/Mar/21

	$\int_{0}^{1} \frac{l n (x + \sqrt{1 - x^{2}})}{x} d x = \int_{0}^{π / 2} \frac{l n (\sin x + \cos x)}{\tan} d x$ $= \int_{0}^{π / 2} \frac{l n (1 + \tan x)}{\tan x} d x + \int_{0}^{π / 2} \frac{l n \cos x}{\tan x} d x$ $= \int_{0}^{\infty} \frac{\ln (1 + x)}{x (1 + x^{2})} d x + \int_{0}^{1} \frac{y l n y}{1 - y^{2}} d y$ $= \int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x + \int_{1}^{\infty} \frac{l n (1 + x)}{x (1 + x^{2})} d x + \underset{n = 0}{\sum^{\infty}} \int_{0}^{1} y^{2 n + 1} l n y d y$ $= \int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x + \int_{0}^{1} \frac{x l n (1 + x) - x l n x}{1 + x^{2}} d x - \underset{n = 0}{\sum^{\infty}} \frac{1}{4 {(n + 1)}^{2}}$ $= \int_{0}^{1} \frac{l n (1 + x)}{x (1 + x^{2})} d x + \int_{0}^{1} [\frac{1}{x} (1 - \frac{1}{(1 + x^{2})})] l n (1 + x) - \frac{x l n x}{1 + x^{2}} d x - \frac{π^{2}}{24}$ $= \int_{0}^{1} \frac{l n (1 + x)}{x} d x - \underset{n = 0}{\sum^{\infty}} {(- 1)}^{n} \int_{0}^{1} x^{2 n + 1} l n x d x - \frac{π^{2}}{24}$ $= \frac{π^{2}}{24} + \underset{n = 0}{\sum^{\infty}} \frac{{(- 1)}^{n}}{4 {(n + 1)}^{2}} = \frac{π^{2}}{24} + (1 - 2^{1 - 2}) \frac{π^{2}}{24} = \frac{π^{2}}{16}$
	Commented by mnjuly1970 last updated on 26/Mar/21

	$v e r y n i c e . . g r a t e f u l . . .$
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