Evaluating $\int_{0}^{x} e^t \sqrt{2 + \sin(2t)} \, dt$ I was recently asked to evaluate the following integral:
$$\int_0^x e^t \sqrt{2 + \sin(2t)} \, dt$$
It was beyond the ken of WolframAlpha, which I find quite discouraging.
Does anyone have an idea of how to tackle this problem?
 A: If Mathematica (which is what Wolfram Alpha runs on) can't find a (closed-form) solution, it's likely it has none, neither in terms of "elementary" functions or even commonly used non-elementary special functions (of which Mathematica has quite a few). There are many integrals like that. It's not certain, but likely. There is a theorem of Liouville that can be applied to determine if elementary solutions are possible, not sure about solutions via more sophisticated functions.
An amazing fact is that even the seemingly very innocent-looking
$\int x^x dx$
has no solution not only in terms of elementary functions, but common non-elementary special functions as well (at least, I haven't seen any, and Mma doesn't seem to know of any).
Failing a closed form, you could try a series expansion, but I suspect it'll be a mess in this case.
If you need a value for a particular $x$, you can approximate it to arbitrary precision via a numerical integration ("quadrature") technique.
EDIT: NEW!!!
Following up on the "series expansion" idea, I found this infinite series expansion. It is based on the idea of representing sine by its complex exponential representation and using the Taylor series for the square root.
$$\int_{0}^{x} e^t \sqrt{2 + \sin(2t)} dt = \sqrt{2} \left(\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^{n+k} (2n)!}{(1 - 2n)(n!) 16^n i^n k! (n-k)! (2i(n - 2k) + 1)} e^{(2i(n - 2k) + 1)x}\right) - \sqrt{2} \left(\sum_{n=0}^{\infty} \sum_{k=0}^{n} \frac{(-1)^{n+k} (2n)!}{(1 - 2n)(n!) 16^n i^n k! (n-k)! (2i(n - 2k) + 1)}\right)$$.
The thing is, this series looks very hypergeometric. The only trouble I see is the presence of the factors $(n-k)!$ and $(2i(n - 2k) + 1)$ in the denominator, which depend on BOTH indices...
A: Ryan Budney's strategy can be expanded upon by WolframAlpha.  It will also provide numerical approximations if asked.  One can also see that the derivative gets twitchy at least twice, which may benefit an explanation of the lack of any representation of the anti-derivative in terms of elementary functions.
