Look up Chebyshev approximation. For example at http://en.wikipedia.org/wiki/Chebyshev_approximation#Chebyshev_approximation.
It's purpose is to solve exactly this sort of problem.
To compute the best approximation, you need to use something like the Remez algorithm. If you're willing to accept something somewhat less than optimal, interpolation at the zeros of Chebyshev polynomials gives a very good result with a lot less effort.
There is a Matlab add-on called Chebfun that specializes in computing high-degree polynomial approximations. See http://www2.maths.ox.ac.uk/chebfun/.
Using the series expansion you suggested is not likely to work very well. It will give you an approximation that is very good when $x$ is close to zero, but very poor when $x$ is large. Looks like you want an approximation that is equally good (in some sense) over the entire interval. This is what the Chebyshev approach gives you.
Reading your question again, it seems that you are involved with signal processing. Design of filters in DSP is often based on Chebyshev approximations. The signal processing guys refer to the Remez algorithm as the "Parks-McClellan" method.
Interpolating at equally-spaced points is definitely not the right thing to do. A famous example due to Runge shows how badly things can go wrong (and actually, its shape is quite similar to your function). You need the points to be more densely spaced at the ends of your interval, and more widely spaced in the middle. A common choice is the set of zeros of a Chebyshev polynomial. For degree $n$, these points are:
$$x_n = \cos\left( \frac{(2k-1)\pi}{2n} \right) \quad (k = 1,2,\ldots,n)$$
These points are suitable for use on the interval $[-1,1]$. You have to shift/scale them to use them on different intervals. See http://en.wikipedia.org/wiki/Chebyshev_nodes for more details.
See also:
http://mathdl.maa.org/images/upload_library/4/vol6/Sarra/Chebyshev.html
A fundamental difference between differentiation and integration is that differentiation is performed at specific points, while integration is done over a region. So while you are correct that a function cannot be differentiated at "sharp points" (more formally, points where the function is not differentiable), there is nothing to stop you from integrating over sharp points (so long as your function is integrable on whatever domain you're integrating over).
Though harder to actually compute, integration is, in a sense, nicer than differentiation. A differentiable function is necessarily continuous, but this is not the case for integrable functions - in fact, any continuous function is integrable. So even if you do have "sharp points", so long as your function is continuous (and even if it's not in pretty much every case you'll ever encounter in high school), you can integrate over any points at which your function isn't differentiable.
Practically, if your domain of integration includes the point $x=1$, then that just means you'll need to split up your integral. For example:
$$\int_0^2|1-x|\,dx=\int_0^1(1-x)\,dx+\int_1^2(x-1)\,dx.$$
Best Answer
Let $y=A\sin\left(2\pi\dfrac xP\right)$.
The arc length is given by the integral
$$\int_{x_0}^{x_1}\sqrt{1+y'^2(x)}dx=\int_{x_0}^{x_1}\sqrt{1+\frac{4\pi^2A^2}{P^2}\cos^2\left(2\pi\dfrac{x}P\right)}dx.$$
The latter has no closed form with usual functions and requires the so-called elliptic integrals.
These functions are called complete or incomplete, depending on whether you cover whole periods of the sinusoid or not.
If you can tolerate an approximation, you could replace the sinusoid by arcs of hyperbolic cosine, which lead to an analytic solution.
$$y=\cosh(x)\to\int \sqrt{1+\sinh^2(x)}dx=\int\cosh(x)\,dx=\sinh(x)+C.$$