# Indefinite Integral

Discussion in 'Homework' started by JustNrik, Jul 12, 2015.

1. ### JustNrikWell-Known Member

$\dpi{100} \bg_white \fn_jvn \large \int e^{sin(x)}dx$

Can someone solve it?

It's not "secxe^sinx + c" just in case.

2. ### Siraraz.Well-Known Member

I tried. I failed.

4. ### EutychiusModerator

It's an indefinite integral so there isn't really a scale to reduce the polynomial down to a finite expression.

If you tried that, you would integrate the first 2-3 parts of it at most which of course a very rough estimation.

But I guess it's the only way because I can't seem to find any other way.

I tried converting the sinx into a complex number expression and use Euler, but it only got more convoluted.

5. ### TheImperialWell-Known Member

Are you sure you can solve it in elementary functions?

Would be much easier if it was finite.

Anyway, you can write e^sinx as Taylor sum and after that integrate (sinx)^n and reverse the sum. [sarcasm] Pretty easy [/sarcasm]

6. ### BlarrgWell-Known Member

You can't solve this using integral from 0 to infinite which I think is what you attempted.

Posters above are indeed correct, you would solve using a series.

e^x = 1 + x + x^2/2! + x^3/3! + x^4/4!...

To get e^sin(x) just replace all the x's above with sin(x)'s taylor series (I don't remember it).

Then you integrate that huge mess and hope you're still alive

EDIT: In terms of an answer. Its super easy. sin(x) has no convergence as it goes to infinity, so e^sin(x) could not possibly converge.

Last edited: Jul 18, 2015
7. ### TheImperialWell-Known Member

It's Indefinite integral, not improper one with ±inf limits.

8. ### BlarrgWell-Known Member

Its been too long since my calc days n2long:

9. ### Heis3nbergWell-Known Member

Isn't there a formula for the integral of sin^n x for general integer x? (The derivation involves integrating by parts to get a recurrence formula). If so, just expand into taylor series like blarrg suggested and apply the general solution. You get the final answer in the form of an infinite series, which may or may not be simplfiable.

10. ### JustNrikWell-Known Member

There is, but it has nothing to do with e^sinx.

11. ### Heis3nbergWell-Known Member

It does if you expand e^sin x into its maclaurin series?

12. ### JustNrikWell-Known Member

You'll get nothing thru integrating the McLaurin serie of e^sinx, it is a indefinite integral. If it was finite it would be easier (in comparison, but it's also a mess), but it isn't.

13. ### Heis3nbergWell-Known Member

Couldnt you write the maclaurin expansion in infinite series notation? Then you get something like

int of sum (sin^n (x) over n!) from n = 0 to inf. dx

Then you reverse summation and integration, and sub in the general formula for the integral of sin^n x, accounting for the even n and odd n cases seperately. You'll get a messy infinite nested sum or a sum of products (can't remember what the integral of sin^n x was). That's... somewhat of an answer right? Since there aren't any answers in terms of elementary functions anyway that's the best I could think of.

Do you have an answer in mind for this question?

14. ### JustNrikWell-Known Member

I don't have mind for this, I'm on vacation. But it's gonna be a great mess to solve it with power series.

15. ### 3.14159Well-Known Member

This kind of integral doesn't have a nice form. Basically, it can't be integrated without using power series or some numerical approximation.