Integrating Trigonometric Integral: Simplifying e^sinx Expression

Click For Summary

Homework Help Overview

The discussion revolves around the integration of the expression y = ∫ e^{sin x} dx, exploring its complexity and potential simplifications. Participants are examining the challenges associated with integrating this trigonometric integral.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss various attempts to simplify the integral, including rewriting it in different forms. There are questions about the feasibility of finding a solution in terms of standard mathematical functions.

Discussion Status

The discussion is ongoing, with participants sharing their experiences of using computational tools like WolframAlpha, which have not yielded satisfactory results. There is a general sense of skepticism regarding the possibility of evaluating the indefinite integral.

Contextual Notes

Some participants note that the integral appears to be intractable, and there is a mention of similar forms that also seem difficult to integrate. The discussion reflects a lack of consensus on the potential for simplification or resolution.

psholtz
Messages
133
Reaction score
0

Homework Statement



I'm trying to integrate the following form:

[tex]y = \int e^{\sin x} dx[/tex]

The Attempt at a Solution



I thought about trying to write something like:

[tex]y = \int e^{\frac{i}{2}e^{-ix} - \frac{i}{2}e^{ix}} dx[/tex]

But this seems to lead down the road of trying to integrate the form

[tex]\int e^{e^x} dx[/tex]

which seems similarly intractable.

Is there a way to reduce the expression to something simpler, or are you just left w/ leaving the expression in a form like:

[tex]y(x) = \int_{x_0}^x e^{\sin t} dt[/tex]
 
Physics news on Phys.org
Yes, I tried Wolfram before posting as well and it came up empty for me too..

My guess is that the integral:

[tex]y = \int e^{\sin x + 2x} dx[/tex]

is just as intractable as the first one, yes?

Thanks for your help..
 
Again, since wolframalpha can't find a solution in terms of standard mathematical functions, I doubt that you'll be able to evaluate the indefinite integral.
 

Similar threads

Replies
1
Views
1K
Replies
7
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
3
Views
2K
Replies
2
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
11K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 19 ·
Replies
19
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K