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shubhamsrg
Group Title
integral [ log( sqrt(1+x) + sqrt(1x) ) dx] from x=0 to 1
 one year ago
 one year ago
shubhamsrg Group Title
integral [ log( sqrt(1+x) + sqrt(1x) ) dx] from x=0 to 1
 one year ago
 one year ago

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

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.0
yep i think so
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yep.. and a dx there,,ofcorse..
 one year ago

vikrantg4 Group TitleBest ResponseYou've already chosen the best response.0
ah.. leave the typos.. concentrate on question. lol
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
simple substitution of x=cos 2t
 one year ago

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

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
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 ?
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
is that \(\ln\)
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
yep..ln..
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i need to reconsider
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
man this is interesting
 one year ago

mukushla Group TitleBest ResponseYou've already chosen the best response.0
let\[t=\ln(\sqrt{1+x}+\sqrt{1x})\]
 one year ago

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

mukushla Group TitleBest ResponseYou've already chosen the best response.0
i think there is
 one year ago

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

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

mukushla Group TitleBest 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
 one year ago

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

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

hartnn Group TitleBest ResponseYou've already chosen the best response.2
lol,i did same exact thing :P
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
i guess,thats integration by parts........???
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
taking u=ln(....),v=1
 one year ago

shubhamsrg Group TitleBest ResponseYou've already chosen the best response.0
bingo,,got it by parts! @hartnn thanks a lot!
 one year ago

hartnn Group TitleBest ResponseYou've already chosen the best response.2
belated welcome :)
 one year ago
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