Showing a hyperbolic cosine of W is a martingale

Let $W$ be standard Brownian motion and $\lambda$ a constant. Prove that $X_t=e^{-\lambda^2t/2}\cosh(\lambda W_t)$ is a martingale.

It is the average of two exponential martingales. Since $\cosh y=\tfrac12(e^y+e^{-y})$, $X_t=\tfrac12\big(e^{\lambda W_t-\lambda^2t/2}+e^{-\lambda W_t-\lambda^2t/2}\big)$, and each term is a martingale. Equivalently, for $f(t,w)=e^{-\lambda^2t/2}\cosh(\lambda w)$ one checks $f_t+\tfrac12f_{ww}=0$, so Ito's formula has no $dt$ term (and integrability is fine).