Calculate $\int_{-\infty}^{\infty} \cos(a t) e^{-\frac{t^2}{2}}dt$ using $E[U_t]$ where $U_t = cos(\sigma W_t)$ where $W_t$ is Brownian Motion

Question

Related: calculate $\int_{-\infty}^{+\infty} \cos(at) e^{-bt^2} dt$

Consider the Brownian Motion $U_t = \cos(\sigma W_t)$, where $W_t$ is a given Brownian Motion. Find $dU_t$ and then $\mathbb{E}[U_t]$. Hence, calcuate $\int_{-\infty}^{\infty} \cos(a t) e^{-\frac{t^2}{2}}dt.$

Using Ito's formula, I've found:

$$E[e^{\frac12\sigma^2t}U_t] = E[X_t] = X_0=1.$$

or

$$E[U_t] = e^{-\frac12\sigma^2t}.$$

(more details here: Using Ito Calculus to find $\mathbb{E}[U_t]$ if $U_t= \cos(\sigma W_t)$ where $W_t$ is Brownian Motion)

From this, I'm a bit cautious on making the jump to solve the integral $\int_{-\infty}^{\infty} \cos(a t) e^{-\frac{t^2}{2}}dt$? How would you do it?