As far as I can tell, it's not trivial, but can be derived by repeatedly exploiting symmetries of the trig functions and properties of logarithms.

One observation is that by symmetry

int_0^pi ln[1 - cos(x)] dx = int 0^pi ln[1 + cos(x)] dx

and so the calculation is equivalent to

1/2 int_0^pi ln[ sin(x)^2 ] dx = 2 int_0^pi/2 ln[ sin(x) ] dx.

Now we can use a similar trick again:

int_0^pi/2 ln[ sin(x) ] dx = int_0^pi/2 ln[ cos(x) ] dx

so

4 int_0^pi/2 ln[ sin(x) ] dx = 2 int_0^pi/2 ln[ sin(x) cos(x) ]

= 2 int_0^pi/2 ln[1/2] dx + 2 int_0^pi/2 ln[sin(2x)] dx

= - pi ln[2] + int_0^pi ln[sin(u)] du

(using substitution u = 2x)

= -pi ln[2] + 2 int_0^pi/2 ln[ sin(x) ] dx

and so the original integral is -pi ln[2].