For $ \int_{-\pi/3}^{\pi/3} \ln(\sin(x)) $ dx, I tried using the Even-Odd method
Even part = $\frac{\ln(\sin(x)) + \ln(\sin(- x))}{2}$ = $\frac{\ln(\sin(x)) + \ln(-1) + \ln(\sin(x))}{2}$ = $\frac{\ln(-1)}{2} + \ln(\sin(x))$
The integral of the even part would be equal to that of the original function
$$\int_{-\pi/3}^{\pi/3} \ln(\sin(x)) dx = \int_{-\pi/3}^{\pi/3} \ln(\sin(x)) + \int_{-\pi/3}^{\pi/3} \frac{1}{2}\ln(-1) dx $$
$$\rightarrow$$ $$0 = \int_{-\pi/3}^{\pi/3} \frac{1}{2}\ln(-1) dx $$
Which is not true.
So when using the Even-Odd method, would you count complex values of the function as being equal to 0? Or is there another explanation?
Thanks