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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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