Sizing a trace on a PCB to carry 2.5 amps
After doing a quick google of "PCB Current Calculator", I found a PCB Current Calculator based on IPC-2152. It bases the width of the track on how much of a temperature rise the trace is allowed to have. It's nice in that it shows how much power you waste through your trace. I would design for your worst-case RMS current, since it's going to be a periodic signal.

If you use 2 oz/ft2 copper instead of the standard 1 oz/ft2 copper, you won't need as wide of a trace to achieve the same resistance. For example, allowing for a 10 oC rise, you can get away with these numbers at 3 A with no copper plane nearby:


177 mil (4.50 mm) on 1/2 oz/ft2 copper
89 mil (2.26 mm) on 1 oz/ft2 (35µm) copper
47 mil (1.19 mm) on 2 oz/ft2 (70µm) copper




Note: IPC-2221 (The standard used in the original answer) uses old measured values for its design charts, and these charts are implemented in many calculators. As best as I can tell, this data was claimed to be 50 years old, which makes IPC-ML-910 (1968) a possible source. As @AlcubierreDrive pointed out, a new standard, IPC-2152, contains new measured data, and presumably is more accurate. More importantly, a comparison of IPC-2221 values gives the following result for trace widths: IPC-2221 (internal) > IPC-2152 > IPC-2221 (external). Actual numbers for the example above (1oz copper) are

IPC-2152:             89 mil 
IPC-2221 (internal): 143 mil (+60%)
IPC-2221 (external):  55 mil (-38%)


Also note that the original numbers in this answer were based on the IPC-2221 internal calculations, which will provide a conservative estimate for all values.