I tried to find an integral of the following,$\DeclareMathOperator{\erfc}{erfc}$

$\int\limits_0^{2\pi} \erfc(a + b\cos(\theta))\erfc(c + d\sin(\theta))\,d\theta $

Where, $a,b,c,d \in \Bbb R$

Now, $\cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2};\quad\sin(\theta) = \frac{e^{j\theta } - e^{-j\theta }}{2j}$

Let $z = {e^{j\theta }}$

Therefore, I can rewrite the integral as, $\oint\limits_{|z| = 1} {\erfc\left( {a + b\left( {\frac{{z + {z^*}}}{2}} \right)} \right)erfc\left( {c + d\left( {\frac{{z - {z^*}}}{{2j}}} \right)} \right)\frac{{dz}}{{iz}}} $

Again, let $f(z) = \erfc\left(a + b\left(\frac{z + z^*}{2}\right)\right)\erfc\left(c + d\left(\frac{z - z^*}{2j}\right) \right)$

Hence , the final integral can be written as,

$\oint\limits_{|z| = 1}f(z)\frac{dz}{iz} = 2\pi f(0) = 2\pi\erfc(a)\erfc(c)$

As $0$ falls inside $|z| = 1$.

Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.