Is this the correct way to calculate IGBT power loss?
Fundamentally yes but not quite 

You have captured the basic concept 

\$ P_{total} = P_{cond} + P_{sw} + P_{leak} \approx P_{cond} + P_{sw} \$

I have added in leakage losses for completeness but this only becomes of significance one you get to series stacked devices.

Conduction loss
This is the subtle difference. Conceptually it is \$ P_{cond} = V_{ce} \cdot I_c\$ however, \$V_{ce}\$ might not be good enough.  This needs to be the saturation voltage at your operating point. Now you may have read the datasheet and chosen \$V_{ce} = 2.1V\$ when \$I_c\$ equals 130A but there is no named part to cross-reference. 

Assuming you have done this correctly then yes \$ P_{cond} = V_{ce} \cdot I_c\$ scaled by duty.

However, there is a more flexible method and that is approximating the IGBT operating point as a series connected DC source \$V_{ce0}\$ and a collector-emitter resistance \$r_c\$



In this example I an interested in an operating point around 20A. 

1V = 18.942mm
zero @ 17.071mm
therefore Vce0 = 0.901V
dI = 20A
dv = 12.527mm = 0.661V
therefore rc = 0.00331R

The instantanious power eqates to

\$P_{cont}(t) = V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t)\$ 

The average loss is therefore 

\$P_{cont}(t) = \frac{1}{T_{sw}}\int_{0}^{T_{sw}}  V_{ce0}(t)\cdot I_c(t) + r_c\cdot i_c^2(t) dt = V_{ce0}\cdot I_{c,avg} + r_c\cdot I_{c,rms}^2\$ 

You can then derive the average and rms currents based upon your waveform. 

A similar process can be done with the diode IF appropriate. NOTE: thermally the co-packaged diode typically becomes the limiting factor in a half-bridge 

Switching loss
AS you have captured... Switching loss is the accumulation of switching energy with respect to the switching frequency 

\$P_{sw} = (E_{on} + E_{off})\cdot f_{sw} \$ 

However... the Eon and Eoff are typically stated at a given operating point 



This means you need to scale the switching energy with respect to your operating point AND the test point. a linear rescaling is usually good enough BUT please check the Eon,off curves to see if a linear interpolation is good enough 

\$P_{sw} = (E_{on,t} + E_{off,t})\cdot \frac{V_{DC}}{V_{DC,test}}\cdot \frac{I_{peak}}{I_{I,test}}\cdot f_{sw} \$

Likewise there is is something comparable with the diode as you accumulate the reverse recovery charge.

The real fun is deriving the average and rms current waveforms and there are given equations for the profile all related to modulation depth