find ∫_(−1) ^1 ln((√(1+x))+(√(1−x)))dx


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Question Number 71314 by mathmax by abdo last updated on 13/Oct/19

$f i n d \int_{- 1}^{1} l n (\sqrt{1 + x} + \sqrt{1 - x}) d x$
Commented by mathmax by abdo last updated on 13/Oct/19
$let I=∫_(−1) ^1 ln((√(1+x))+(√(1−x)))dx ⇒I=2∫_0 ^1 ln((√(1+x))+(√(1−x)))dx (function even) I=_(x=cos(2t)) 2∫_(π/4) ^0 ln((√(1+cos(2t)))+(√(1−cos(2t))))(−2)sin(2t)dt =4 ∫_0 ^(π/4) ln((√2)cost +(√2)sint)sin(2t)dt =4∫_0 ^(π/4) ln((√2)((√2)cos(t−(π/4)))sin(2t)dt =4 ∫_0 ^(π/4) {ln(2)+ln(cos((π/4)−t)}sin(2t)dt =4ln(2)∫_0 ^(π/4) sin(2t)dt +4 ∫_0 ^(π/4) ln(cos((π/4)−t))sin(2t)dt =4ln(2)[−(1/2)cos(2t)]_0 ^(π/4) +4 ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt =2ln(2) +4 ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt changement (π/4)−t=u givd ∫_0 ^(π/4) sin(2t)ln(cos((π/4)−t))dt =−∫_0 ^(π/4) sin(2((π/4)−u))ln(cosu)(−du) =∫_0 ^(π/4) sin((π/2)−2u)ln(cosu)du =∫_0 ^(π/4) cos(2u) ln(cosu)du by parts f^′ =cos(2u) and g =ln(cosu) ⇎ ∫_0 ^(π/4) cos(2u)ln(cosu)du =[(1/2)sin(2u)ln(cosu)]_0 ^(π/4) −∫_0 ^(π/4) (1/2)sin(2u)(((−sinu)/(cosu)))du=(1/2)ln((1/(√2)))+(1/2) ∫_0 ^(π/4) ((sin(u)(2sinu cosu))/(cosu))du =−(1/2)ln((√2)) +∫_0 ^(π/4) sin^2 u du =−(1/4)ln(2)+(1/2)∫_0 ^(π/4) (1−cos(2u))du =−(1/4)ln(2)+(π/8)−(1/4)[sin(2u)]_0 ^(π/4) =−(1/4)ln(2)+(π/8)−(1/4) ⇒ I =2ln(2)−ln(2)+(π/2) −1 =ln(2)+(π/2) −1$
$l e t I = \int_{- 1}^{1} l n (\sqrt{1 + x} + \sqrt{1 - x}) d x \Rightarrow I = 2 \int_{0}^{1} l n (\sqrt{1 + x} + \sqrt{1 - x}) d x$ $(f u n c t i o n e v e n)$ $I =_{x = c o s (2 t)} 2 \int_{\frac{π}{4}}^{0} l n (\sqrt{1 + c o s (2 t)} + \sqrt{1 - c o s (2 t)}) (- 2) s i n (2 t) d t$ $= 4 \int_{0}^{\frac{π}{4}} l n (\sqrt{2} c o s t + \sqrt{2} s i n t) s i n (2 t) d t$ $= 4 \int_{0}^{\frac{π}{4}} l n (\sqrt{2} (\sqrt{2} c o s (t - \frac{π}{4})) s i n (2 t) d t$ $= 4 \int_{0}^{\frac{π}{4}} {l n (2) + l n (c o s (\frac{π}{4} - t)} s i n (2 t) d t$ $= 4 l n (2) \int_{0}^{\frac{π}{4}} s i n (2 t) d t + 4 \int_{0}^{\frac{π}{4}} l n (c o s (\frac{π}{4} - t)) s i n (2 t) d t$ $= 4 l n (2) {[- \frac{1}{2} c o s (2 t)]}_{0}^{\frac{π}{4}} + 4 \int_{0}^{\frac{π}{4}} s i n (2 t) l n (c o s (\frac{π}{4} - t)) d t$ $= 2 l n (2) + 4 \int_{0}^{\frac{π}{4}} s i n (2 t) l n (c o s (\frac{π}{4} - t)) d t c h a n g e m e n t \frac{π}{4} - t = u g i v d$ $\int_{0}^{\frac{π}{4}} s i n (2 t) l n (c o s (\frac{π}{4} - t)) d t = - \int_{0}^{\frac{π}{4}} s i n (2 (\frac{π}{4} - u)) l n (c o s u) (- d u)$ $= \int_{0}^{\frac{π}{4}} s i n (\frac{π}{2} - 2 u) l n (c o s u) d u = \int_{0}^{\frac{π}{4}} c o s (2 u) l n (c o s u) d u$ $b y p a r t s f^{^{'}} = c o s (2 u) a n d g = l n (c o s u) ⇎$ $\int_{0}^{\frac{π}{4}} c o s (2 u) l n (c o s u) d u = {[\frac{1}{2} s i n (2 u) l n (c o s u)]}_{0}^{\frac{π}{4}}$ $- \int_{0}^{\frac{π}{4}} \frac{1}{2} s i n (2 u) (\frac{- s i n u}{c o s u}) d u = \frac{1}{2} l n (\frac{1}{\sqrt{2}}) + \frac{1}{2} \int_{0}^{\frac{π}{4}} \frac{s i n (u) (2 s i n u c o s u)}{c o s u} d u$ $= - \frac{1}{2} l n (\sqrt{2}) + \int_{0}^{\frac{π}{4}} {s i n}^{2} u d u$ $= - \frac{1}{4} l n (2) + \frac{1}{2} \int_{0}^{\frac{π}{4}} (1 - c o s (2 u)) d u$ $= - \frac{1}{4} l n (2) + \frac{π}{8} - \frac{1}{4} {[s i n (2 u)]}_{0}^{\frac{π}{4}}$ $= - \frac{1}{4} l n (2) + \frac{π}{8} - \frac{1}{4} \Rightarrow$ $I = 2 l n (2) - l n (2) + \frac{π}{2} - 1 = l n (2) + \frac{π}{2} - 1$
	Answered by mind is power last updated on 13/Oct/19

	$x = \cos (2 a)$ $\Rightarrow dx = - 2 \sin (2 a)$ $1 + x = {2 \cos}^{2} (a) : 1 - x = {2 \sin}^{2} (a)$ $\int_{- 1}^{1} \ln (\sqrt{1 + x} + \sqrt{1 - x}) dx = 2 \int_{0}^{\frac{π}{2}} \ln (\sin (a) \sqrt{2} + \cos (a) \sqrt{2}) \sin (2 a) da$ $\int_{0}^{\frac{π}{2}} \ln (\sqrt{2} \sin (a) + \sqrt{2} \cos (a)) \sin (2 a) da = [\frac{- \cos (2 a)}{2} \ln (\sqrt{2} \sin (a) + \sqrt{2} \cos (a)) \int \frac{\cos (2 a)}{2} . \frac{\cos (a) - \sin (a)}{\sin (a) + \cos (a)} da$ $= \ln (\sqrt{2)} + \frac{1}{2} \int {(\cos (a) - \sin (a))}^{2} da$ $= \ln \sqrt{2} + \frac{1}{2} \int 1 - \sin (2 a) da$ $= \ln (\sqrt{2}) + \frac{1}{2} (\frac{π}{2}) - \frac{1}{2}$ $I = 2 \ln (\sqrt{2}) + \frac{π}{2} - 1$
	Commented by MJS last updated on 13/Oct/19

	$my solution is wrong (why ?)$ $but yours also$ $the integral is approximately 1.26394$
	Commented by mind is power last updated on 13/Oct/19

	$my answer * 2$
	Commented by mathmax by abdo last updated on 13/Oct/19

	$t h a n k s s i r .$
	Commented by MJS last updated on 13/Oct/19

	$. . . and I had a sign error . . .$
	Answered by MJS last updated on 13/Oct/19

	$I = \int \ln (\sqrt{1 + x} + \sqrt{1 - x}) d x =$ $by parts$ $\int u^{'} v = u v - \int {u v}^{'}$ $u^{'} = 1 \to u = x$ $v = \ln (\sqrt{1 + x} + \sqrt{1 - x}) \to v^{'} = \frac{1 - x^{2} - \sqrt{1 - x^{2}}}{2 x (1 - x^{2})}$ $= x \ln (\sqrt{1 + x} + \sqrt{1 - x}) - \frac{1}{2} \int \frac{1 - x^{2} - \sqrt{1 - x^{2}}}{1 - x^{2}} d x$ $- \frac{1}{2} \int \frac{1 - x^{2} - \sqrt{1 - x^{2}}}{1 - x^{2}} d x =$ $[t = \arcsin x \to d x = \sqrt{1 - x^{2}} d t]$ $= \frac{1}{2} \int \cos t d t - \frac{1}{2} \int d t =$ $= \frac{1}{2} \sin t - \frac{t}{2} = - \frac{x}{2} + \frac{1}{2} \arcsin x$ $\Rightarrow I = x \ln (\sqrt{1 + x} + \sqrt{1 - x}) - \frac{x}{2} + \frac{1}{2} \arcsin x + C$
	Commented by MJS last updated on 13/Oct/19

	$\underset{- 1}{\int^{1}} \ln (\sqrt{1 + x} + \sqrt{1 - x}) d x = \frac{π}{2} - 1 + \ln 2$
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