Leading behaviour of given integral

Question

I am trying to evaluate the leading behaviour of $$\int_{-1}^1\sin[x(t-\sin t)] \sinh t \,dt$$ as $x$ tends to infinity.
As $\sin[x(t-\sin t)] = \Im [ e^{ixt}e^{-ix\sin t }]$
$\int_{-1}^1\sin[x(t-\sin t)] \sinh t \,dt = \int_{-1}^1 \Im[ e^{ix(t-sint )}]\sinh t dt$
Here $t=0$ is a stationary point of $t-\sin t$, but $\sinh t$ vanishes at $t=0$. How to proceed with this using method of stationary phase?