How to convert AC to DC
1000 µF at this voltage isn't terribly big.  Are you limited by size or something?
To completely get rid of the ripple and produce 5 V, you need to add a voltage regulator after the capacitor.
12 VRMS = 17 VPeak, which, minus the two diode drops, is the peak DC voltage you'll see at the output of the rectifiers: 17 - 1.1 - 1.1 = 14.8 V.  So there's no threat of exceeding the input limits of the regulator (35 V input).
If the ripple is 8.3 V, then the DC voltage will be varying from 6.5 V to 15 V.  This is just barely high enough to feed into the regulator without dropping out of regulation, since the 7805 has about 1.5 V dropout at 1 A (depending on temperature).  So yes, you should use a slightly higher capacitor (or multiple capacitors in parallel, if space is an issue).
 (Source: Alan Marshall)
Here's a guide to each stage of the power supply circuit.
Also:
Real life power line voltages vary from one outlet to the next, and the frequency varies by country.  You need to calculate the low line/high load condition to make sure it doesn't drop below regulation, as well as the high line/low load condition to make sure it doesn't exceed the regulator's input voltage limit.  These are the generally recommended values:

JP:  85 VAC to 110 VAC (+10%, -15%), 50 and 60 Hz
US: 105 VAC to 132 VAC (+10%), 60 Hz
EU: 215 VAC to 264 VAC (+10%), 50 Hz