integral from 0 to 1 of [e^x(cos(e^x))]dx ?

Question

jennychan12 · Answer

$$\int\limits_{0}^{1} [e^x \cos (e^x)]dx$$

kinggeorge · Answer

Do a u-substitution with $u=\sin(e^x)$, and your solution should pop right out.

hartnn · Answer

did you mean u=e^x ?

jennychan12 · Answer

but if you do that, then du = e^xcos(e^x) and there's no cos (e^x) in the question.

hartnn · Answer

try u=e^x

jennychan12 · Answer

that'd just be ucosu

kinggeorge · Answer

$u=\sin(e^x)\implies du=e^x\cos(e^x)dx$. So your integral becomes$$\int du=u$$

kinggeorge · Answer

Substitute back for $u$, and you get $$\int_0^1 e^x\cos(e^x)dx=\sin(e^x)|_0^1$$

hartnn · Answer

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 · Answer

oh wait. my bad. i thought u said u = sin u
ok i see now

kinggeorge · Answer

Yup, I like to call my method "guessing the solution and proving you're right"

kinggeorge · Answer

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 · Answer

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 · Answer

how do we need integration by parts ?? O.o
u= e^x
du=e^xdx

kinggeorge · Answer

Oh. Right. You don't. Ignore my ramblings.