integration of """e^sinx.dx"""?

Answer 1

Sorry tried a lot, searched a lot & surfed a lot but found no solution, I think Pascal is right.

P.s- Saket it is ∫e^sinx dx & not ∫(e^x)*sin x dx, if it was that, it would have been trivial.

Answer 2

I don't have access to a pen and paper, but intuitively, you should be able to solve it by expanding exp.(sinx) using the e^x series, substituting sin x for x.

You will have an infinite series, for the the general term is (sin x)^n / n! . Integrate the general term and then sum the general term from 0 to infinity. Voila!

Answer 3

Can only be done numerically. That function has no elementary antiderivative.

Edit: and what's with the idiots who don't know the difference between an integral and a derivative?

Answer 4

Ever because of the fact that i replaced into little and til on the instant time i've got consistently called my mum "Mummy" and my dad "Daddy". in basic terms because of the fact they by no skill replied to "mum"/"dad". and that's often caught, i think. same with my siblings, too. I swear every time my mom calls me at artwork, and that i answer saying "hi, mummy", i'm getting the main strangest seems from my artwork colleagues. They consistently discover that hilarious and generally take the mic. and that i can actually see why it relatively is humorous lol yet, it relatively is by no skill replaced, and that i doubt it is going to. mom and pa will consistently be Mummy and Daddy. And in particular because of the fact they gained't respond to something from their babies! : )

Answer 5

Sorry, pal this little @#$ can't be solved without numerical values.

Answer 6

the given integration is e^sinx.dx
we use integrating by parts
I=sinxe^-integ of cosxe^dx
=sinxe^-(cosxe^-integ of cosxe^dx)
=sinxe^-(cosxe^)-I
2I=sinxe^-cosxe^
I=sinxe^-cosxe^/2.

Answer 7

cos x+ C

Answer 8

u = sin x du/dx = cos x
derivative of "" e^u ""= e^u * du/dx --where u is any expression--

therefore the answer is (e^sinx) * cos x