find-1-1-ln-1-x-1-x-dx-

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

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

\dots and I had a sign error \dots

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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