jennychan12
  • jennychan12
integral from 0 to 1 of [e^x(cos(e^x))]dx ?
Mathematics
chestercat
  • chestercat
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jennychan12
  • jennychan12
\[\int\limits_{0}^{1} [e^x \cos (e^x)]dx\]
KingGeorge
  • KingGeorge
Do a u-substitution with \(u=\sin(e^x)\), and your solution should pop right out.
hartnn
  • hartnn
did you mean u=e^x ?

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jennychan12
  • jennychan12
but if you do that, then du = e^xcos(e^x) and there's no cos (e^x) in the question.
hartnn
  • hartnn
try u=e^x
jennychan12
  • jennychan12
that'd just be ucosu
KingGeorge
  • KingGeorge
\(u=\sin(e^x)\implies du=e^x\cos(e^x)dx\). So your integral becomes\[\int du=u\]
KingGeorge
  • KingGeorge
Substitute back for \(u\), and you get \[\int_0^1 e^x\cos(e^x)dx=\sin(e^x)|_0^1\]
hartnn
  • hartnn
u =e^x du = e^x dx \(\int cos udu\) ohh..now i could say that u =sin e^x is a better substitution..
jennychan12
  • jennychan12
oh wait. my bad. i thought u said u = sin u ok i see now
KingGeorge
  • KingGeorge
Yup, I like to call my method "guessing the solution and proving you're right"
KingGeorge
  • KingGeorge
Although if you really had no clue of the solution, \(u=e^x\) would be a fine substitution. You would just have to integrate by parts.
hartnn
  • hartnn
whenever i see a non-standard angle with sin/cos/.. i'll call that as 'bad angle' and put u= bad angle.... example : sin x^2 , cos log x ....
hartnn
  • hartnn
how do we need integration by parts ?? O.o u= e^x du=e^xdx
KingGeorge
  • KingGeorge
Oh. Right. You don't. Ignore my ramblings.

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