$\int_0^{2\pi }\mathrm{erfc}\left( \cos ( {a + \theta } )\right)\mathrm{erfc}\left( {\sin ( {a + \theta } )} \right)d\theta$? I have found a solution of the following integral:
$\int_{0}^{2\pi} \operatorname{erfc}[\cos(a+t)] dt=2\pi$
But I am trying to solve the following integral:
$\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)\operatorname{erfc}\left( {\sin \left( {a + \theta } \right)} \right)d\theta } $
Is there any hint that will help me to solve this type of problem. Thank you.
 A: I'll present a strategy that would also work with your first integral; maybe it's what you did, or similar to same. With $z=e^{i\theta}$, the second problem becomes that of evaluating the anticlockwise contour integral$$\oint_{|z|=1}\operatorname{erfc}\left(\frac{e^{ia}z+e^{-ia}z^\ast}{2}\right)\operatorname{erfc}\left(\frac{e^{ia}z-e^{-ia}z^\ast}{2i}\right)\frac{dz}{iz}.$$Using $\oint f(z)\frac{dz}{iz}=2\pi f(0)$, this is $2\pi\operatorname{erfc}^2(0)=2\pi$. It works.
A: If $g(x)$ is periodic with a period $T$, then $$\int_{k}^{T+k} g(x) dx=\int_{0}^{T} g(x) dx ~~~~~(1)$$
$$I=\int_{0}^{2\pi} \text{Erfc}\cos(a+x)~ \text{Erf}\sin(a+x)] dt=\int_{a}^{2\pi+a} \text{Erfc}\cos(t)~ \text{Erf} \sin(t)~dt.$$
From (1), we get
$$I=\int_{0}^{2\pi} \text{Erfc}\cos(t)~ \text{Erf} \sin(t)~dt~~~~~(2)$$
Next note the property that  $$\int_{0}^{2a} f(x) dx=\int_{0}^{a}[ f(x)x+ f(2a-x)] dx~~~~(3).$$
As $\cos(2\pi-x)=\cos x, \sin(2\pi-x)=-\sin x$ from (3), we get
$$I=\int_{0}^{\pi} [[\text{Erfc}(\sin x) ~\text{Erfc}(\cos x)+[\text{Erfc}(-\sin x) ~\text{Erfc}(\cos x)] dx~~~~(4)$$
Next, use $\text{Erf}(-z)=2-\text{Erf}(z)$,
to get from (4)
$$I=2\int_{0}^{\pi} \text{Erf}(\cos x) dx= I=2\pi$$
Here we have used this integral independeny of $a$, from MSE:
Integral of $\int\limits_0^{2\pi } {\operatorname{erfc}\left( {\cos \left( {a + \theta } \right)} \right)d\theta } $?
