I'm trying to take the following integral. $$I = \int_0^{2\pi}d\theta \ln\left[\sqrt{1 - (\hat{\rho}\cdot\vec{a})^2} + i\hat{\rho}\cdot\vec{a}\right],$$ where $\hat{\rho} = (\cos\theta,\sin\theta)$ is the unit vector in polar coordinates, and $\vec{a} = (a_x,a_y) = \Vert a \Vert(\cos\phi,\sin\phi)$ is some arbitrary vector, and it is known that $a \equiv \Vert\vec{a}\Vert < 1$. We can choose $\phi=0$, such that $$ I = \int^{2\pi}_0 d\theta \ln\left(\sqrt{1 - a^2\cos^2\theta} + ia\cos\theta\right).$$
I have tried change of variables, factorization etc, but get nowhere. Is there any hope of being able to solve it?
$$I=\int_{0}^{2\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx= 2\int_{0}^{\pi} ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]~ dx.$$ Next use $\int_{0}^{a} f(x) dx= \int_{0}^{a} f(a-x) dx$. Then $$2I=2\int_{0}^{\pi}\left( \ln[\sqrt{1-a^2\cos^2 x}+ai \cos x]+\ln[\sqrt{1-a^2\cos^2 x}-ai \cos x]\right)~ dx.$$ $$2I=2\int_{0}^{\pi} \ln[ 1-a^1\cos^2 x+ a^2 \cos^2 x]~dx=2\int_{0}^{\pi} \ln 1 ~dx =0$$ Hence $$I=0.$$