Here's the question you clicked on:
shubhamsrg
integral [ log( sqrt(1+x) + sqrt(1-x) ) dx] from x=0 to 1
\[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})\] ???
yep.. and a dx there,,ofcorse..
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 ?
man this is interesting
let\[t=\ln(\sqrt{1+x}+\sqrt{1-x})\]
is there a standard formula for: \[\int\limits_{}^{}\ln(1+\sqrt{1-x^2})\] if yes,then this is so simple!
\[=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1- x})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1-x^2})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1-x^2})^2 dx\]
\[=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1-x^2}) dx\]
\[\ln(1+\sqrt{1-x^2})=\text{sech}^{-1}x+\ln x\]this is hard to get o.O
http://www.wolframalpha.com/input/?i=%5Cint+%5Cln+%281%2B%5Csqrt%7B1-x%5E2%7D%29+dx
http://www.wolframalpha.com/input/?i=integral+of+ln%281%2Bsqrt%7B1-x%5E2%7D%29&dataset=
lol,i did same exact thing :P
i guess,thats integration by parts........???
bingo,,got it by parts! @hartnn thanks a lot!