how to evaluate the integral: (e^sinx)cosx/(sinx+1)dx
let u=sinx+1 du=cosx dx if u=sinx+1, then u-1=sinx see if this helps
\[\int\limits_{}^{}\frac{e^{u-1}}{u} du\]
\[\frac{1}{e}\int\limits_{}^{}\frac{e^u}{u}du\]
maybe this is the wrong substitution let me look at it some more
well, thanx, I was trying to do like u=1/(sinx+1) dv=(e^sinx)cosx du=-cosx/((sinx+1)^2) v=(e^sinx)
but I still get stuck after this
i dont think this is an elementary integral this is wht i see \[\int\limits\limits_{}^{}\frac{e^{\sin(x)}*\cos(x)}{sinx+1}dx\]
is that right?
yes
I got this (e^sinx)/(sinx+1)+\[\int\limits (e^\sin(x))\cos(x)/(1+sinx)^2=\int\limits e^\sin(x)cosx/(1+sinx)\]
this is an exponential integral
what class is this?
cal2
what? what book?
calculus 202
the book is called calculus 202
its not from the book
this one is too hard for cal 2 i would think i haven't even heard of this exponential interal have you?
no, I spent more than two hours on this
its the first time I get stuck on one question like this
i think this is too hard for cal 2 you should email your teach and tell him its not elementary integral
I have already done that, but he told me this one is super duper easy, and thats why I still trying to do it
Thanx for ur help anyway
you wrote it down correctly?
you have no doubts it was written incorrectly?
yes, I look at it like thousand times
it is not easy
cosx is not in exponent part right?
yes I think I might just leave it
cow might say something in a second let him say something first and then give up
well give up if he doesn't say anything that is positive about integrating this
Ill see wht he say tom.
i was able to write it in the form e^(e^s) and this is what one site said about this http://mathforum.org/library/drmath/view/52023.html
yeah i dont think this has a elementary anti-derivative
and cow is really smart
so i believe in him
and myself i guess since he confirm what i thought lol
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