I am trying to find $\mathbb{E}\left[\cosh \left(B_{t}\right)\right]$ without using Itô's formula. By definition of expected value I know
$$\mathbb{E}\left[\cosh \left(B_{t}\right)\right]=\int_{\mathbb{R}} \cosh(x) f(x) dx$$ where $f(x)$ is the pdf of the Brownian motion $B_t$. As $B_t \sim N(0,t)$ then $f(x)=\frac{1}{\sqrt{t2\pi}}e^{\frac{-(x^2)}{2t}}$. So $$\mathbb{E}\left[\cosh \left(B_{t}\right)\right]=\int_{\mathbb{R}} \cosh(x) \frac{1}{\sqrt{t2\pi}}e^{\frac{-(x^2)}{2t}} dx$$
Im not sure if it's right and how should I proceed.
Recall that $\cosh (x) = \frac{e^x + e^{-x}}{2}$, so that: $$\begin{align*} \mathbb{E}(\cosh (B_t)) &= \frac{1}{2} (\mathbb{E}(e^{B_t}) + \mathbb{E}(e^{-B_t})) \\ & = \mathbb{E}(e^{B_t}) & \text{since } B_t \stackrel{D}{=} -B_t \\ &= \frac{t^2}{2} \end{align*}$$