Question

integral [ log( sqrt(1+x) + sqrt(1-x) ) dx] from x=0 to 1

Answer 1

\[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})\]
???

Answer 2

yep i think so

Answer 3

yep.. and a dx there,,ofcorse..

Answer 4

ah.. leave the typos.. concentrate on question. lol

Answer 5

simple substitution of x=cos 2t

Answer 6

then our
dx = -2sin 2t dt..
dont you think that will create a problem ?
altough the sq. roots will be eliminated..

Answer 7

how'd you get that ?
what i got was inside the integral
(something) + log( cos t + sin t ) sin 2t dt

are you missing the log there buddy ?

Answer 8

is that \(\ln\)

Answer 9

yep..ln..

Answer 10

i need to reconsider

Answer 11

man this is interesting

Answer 12

let\[t=\ln(\sqrt{1+x}+\sqrt{1-x})\]

Answer 13

is there a standard formula for:
\[\int\limits_{}^{}\ln(1+\sqrt{1-x^2})\]
if yes,then this is so simple!

Answer 14

i think there is

Answer 15

\[=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})^2 dx\]
\[=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1-x^2})^2 dx\]
\[=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1-x^2})^2 dx\]

Answer 16

\[=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1-x^2}) dx\]

Answer 17

\[\ln(1+\sqrt{1-x^2})=\text{sech}^{-1}x+\ln x\]this is hard to get o.O

Answer 18

http://www.wolframalpha.com/input/?i=%5Cint%5Climits%5Climits%5Climits+%5Climits_0%5E1%5Cln+%281%2B%5Csqrt%7B1-x%5E2%7D%29+dx&dataset=

Answer 19

http://www.wolframalpha.com/input/?i=%5Cint+%5Cln+%281%2B%5Csqrt%7B1-x%5E2%7D%29+dx

Answer 20

http://www.wolframalpha.com/input/?i=integral+of+ln%281%2Bsqrt%7B1-x%5E2%7D%29&dataset=

Answer 21

lol,i did same exact thing :P

Answer 22

i guess,thats integration by parts........???

Answer 23

taking u=ln(....),v=1

Answer 24

bingo,,got it by parts! @hartnn thanks a lot!

Answer 25

belated welcome :)