Let me do a simple calculation for you , as your inverter is capable of delivering an active power of 500Watt , so I presume that you'll be using a 12V Battery at the input side.So if i calculate the Input current flowing in the primary winding of your ferrite transformer,it'll be 500/12 Ampere i.e 41.66 Ampere . If i presume the Rds =8milliohm of your switching device in ON state, i'll get a power discipation of 13.8 watt (using P= I*I*Rds) which is quite high , and will also need a BIG HEATSINK or Forced air cooling or both to discipate such amount of heat
so definetly you are not gonna get any IC which incorporates a switching device capable of switching a High Current (42 Ampere)..