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

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\[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})\] ???
yep i think so
yep.. and a dx there,,ofcorse..

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ah.. leave the typos.. concentrate on question. lol
simple substitution of x=cos 2t
then our dx = -2sin 2t dt.. dont you think that will create a problem ? altough the sq. roots will be eliminated..
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 ?
is that \(\ln\)
i need to reconsider
man this is interesting
is there a standard formula for: \[\int\limits_{}^{}\ln(1+\sqrt{1-x^2})\] if yes,then this is so simple!
i think there is
\[=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1-x^2})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1-x^2})^2 dx\]
\[=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1-x^2}) dx\]
\[\ln(1+\sqrt{1-x^2})=\text{sech}^{-1}x+\ln x\]this is hard to get o.O
lol,i did same exact thing :P
i guess,thats integration by parts........???
taking u=ln(....),v=1
bingo,,got it by parts! @hartnn thanks a lot!
belated welcome :)

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