Integral including functions $\operatorname{erfc}(.), \exp(.) $ and $ \cos(.)$ I have following integral. MATHEMATICA evaluates it as follows for $a>0$:
$$I=\int_{0}^{\pi/2}\cos (\theta ) e^{a^2 \cos ^2(\theta )} \text{erfc}(a \cos (\theta ))d\theta=\frac{\sqrt{\pi } \left(1-e^{a^2} \text{erfc}\left(a\right)\right)}{2 a}$$
However, I have no clue how this result comes. I have checked few integral table books such as Table of Integrals, Series, and Products, and also functions.wolfram.com. But I could not find matching expressions. 
Does anyone have an idea?
 A: Probably not the most direct way to obtain the result, but one possible method:
From the integral representation DLMF
\begin{equation}
 \int_{0}^{\infty}\frac{e^{-a^2t}}{\sqrt{t+\cos^{2}\theta}}\mathrm{d}t=\frac{\sqrt{\pi}}{a}e^{a^2\cos^{2}\theta}\operatorname{erfc}\left(a\cos\theta\right) \tag{1}\label{1}
\end{equation} 
the integral can be transformed into
\begin{equation}
 I=\frac{a}{\sqrt{\pi}}\int_{0}^{\pi/2}\cos \theta \,d\theta \int_{0}^{\infty}\frac{e^{-a^2t}}{\sqrt{t+\cos^{2}\theta}}\,\mathrm{d}t
\end{equation} 
By changing the integration order,
\begin{align}
 I&=\frac{a}{\sqrt{\pi}}\int_{0}^{\infty}e^{-a^2t}\,dt\int_{0}^{\pi/2}\frac{\cos \theta }{\sqrt{t+1-\sin^2\theta}}\,d\theta\\
 &=\frac{a}{\sqrt{\pi}}\int_{0}^{\infty}e^{-a^2t}\arcsin \frac{1}{\sqrt{t+1}}\,dt
\end{align}
Now, integrating by parts, ($u'=e^{-a^2t},v=\arcsin \frac{1}{\sqrt{t+1}}$), we get
\begin{align}
 I&=\frac{a}{\sqrt{\pi}}\left[\frac{\pi}{2a^2}-\frac{1}{2a^2}\int_{0}^{\infty}\frac{e^{-a^2t}}{\sqrt{t}(t+1)} \right]\\
 &=\frac{a}{\sqrt{\pi}}\left[\frac{\pi}{2a^2}-\frac{1}{a^2}\int_{0}^{\infty}\frac{e^{-a^2u^2}}{u^2+1} \right]
\end{align} 
(by changing $t=u^2$). Using the integral representation DLMF
\begin{equation}
 \operatorname{erfc}(a)=\frac{2}{\pi}e^{-a^{2}}\int_{0}^{\infty}\frac{e^{-a^{2}u^%
{2}}}{u^{2}+1}\mathrm{d}u  \tag{2}\label{2}
\end{equation} 
the final result is obtained:
\begin{equation}
I= \frac{\sqrt{\pi } \left(1-e^{a^2} \operatorname{erfc}\left(a\right)\right)}{2 a}
\end{equation} 
\eqref{1} This identity is retrieved by changing $t=u^2-\cos^2\theta$ in the integral.
\eqref{2} 
With $A=a^2$, defining
\begin{align}
 J(A)&=\frac{2}{\pi}\int_0^\infty\frac{e^{-A\left( u^2+1 \right)}}{u^2+1}\\
 \frac{dJ(A)}{dA}&=-\frac{2}{\pi}\int_0^\infty e^{-A\left( u^2+1 \right)}\,du\\
 &=-\frac{1}{\sqrt{\pi A}}e^{-A}
\end{align} 
and remarking that $J(0)=1$, 
\begin{align}
 J(A)&=1-\frac{1}{\sqrt{\pi}}\int_0^A\frac{e^{-s}}{\sqrt{s}}\,ds\\
 &=1-\operatorname{erf}(a)
\end{align} 
