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shubhamsrg
 4 years ago
integral [ log( sqrt(1+x) + sqrt(1x) ) dx] from x=0 to 1
shubhamsrg
 4 years ago
integral [ log( sqrt(1+x) + sqrt(1x) ) dx] from x=0 to 1

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lgbasallote
 4 years ago
Best ResponseYou've already chosen the best response.0\[\huge \int \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1 x})\] ???

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0yep.. and a dx there,,ofcorse..

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ah.. leave the typos.. concentrate on question. lol

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2simple substitution of x=cos 2t

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0then our dx = 2sin 2t dt.. dont you think that will create a problem ? altough the sq. roots will be eliminated..

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0how'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 ?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0man this is interesting

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0let\[t=\ln(\sqrt{1+x}+\sqrt{1x})\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2is there a standard formula for: \[\int\limits_{}^{}\ln(1+\sqrt{1x^2})\] if yes,then this is so simple!

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2\[=(1/2)\int\limits \limits_0^1 \log (\sqrt{1 + x} + \sqrt{1 x})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \log (2+2\sqrt{1x^2})^2 dx\] \[=(1/2)\int\limits \limits_0^1 \ln 2+\ln (1+\sqrt{1x^2})^2 dx\]

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2\[=(1/2)\int\limits\limits\limits \limits_0^1 \ln 2+\int\limits\limits\limits \limits_0^1\ln (1+\sqrt{1x^2}) dx\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\ln(1+\sqrt{1x^2})=\text{sech}^{1}x+\ln x\]this is hard to get o.O

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=%5Cint+%5Cln+%281%2B%5Csqrt%7B1x%5E2%7D%29+dx

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2http://www.wolframalpha.com/input/?i=integral+of+ln%281%2Bsqrt%7B1x%5E2%7D%29&dataset=

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2lol,i did same exact thing :P

hartnn
 4 years ago
Best ResponseYou've already chosen the best response.2i guess,thats integration by parts........???

shubhamsrg
 4 years ago
Best ResponseYou've already chosen the best response.0bingo,,got it by parts! @hartnn thanks a lot!
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