Expectation of solution to SDE $dX_t=-\tanh(X_t) dt + dW_t$ 
How do I calculate the expectation of the process given by the SDE $$dX_t=-\tanh(X_t) dt + dW_t, \qquad X_0=x_0$$ and $W_t$ a Wiener process?

If I start with 
$$
d\left(e^{t/2}\sinh(X_t)\right) = e^{t/2}\cosh(X_t)dW_t\tag{1}
$$
which gives 
$$
\sinh(X_t) = e^{-t/2}\sinh(x_0)+\int_0^t e^{-(t-s)/2}\cosh(X_s)dW_s\label{a}\tag{2}
$$
then, at first sight, the expectation would seem to be 
$$
\mathbb{E} [\sinh(X_t)\,|\,X_0] = e^{-t/2}\sinh(x_0)\tag{3}
$$
However, it seems that the Itô integral in (\ref{a}) may be not zero in expectation, that it is a local martingale but not a martingale? Perhaps also that 
$$
\mathbb{E} [\sinh(X_t)\,|\,X_0] \le e^{-t/2}\sinh(x_0)\tag{4}\quad ?
$$
In general, how can I calculate the expectation of a stochastic integral such as
$$
\int_0^t e^{-(t-s)} \cosh(X_s) dW_s\ \tag{5}
$$
for a process $\{X_t\}$ such as above? Is there any way to do it explicitly? 
I've been out of the game almost 20yrs, I'm a little rusty, so if what I have posted above is incorrect then please forgive me.
 A: Step 1: Let's first construct a weak solution to the SDE $$dX_t = - \tanh(X_t) \, dt + dB_t, \qquad X_0 = x_0. \tag{1}$$
For a Brownian motion $(W_t)_{t \geq 0}$ on a probability space $(\Omega,\mathcal{A},\mathbb{P})$ we define
$$\tilde{X}_t :=- x_0 + W_t \quad \text{and} \quad \tilde{B}_t := W_t - \int_0^t \tanh(-x_0+W_s) \, ds.$$
Then
$$d\tilde{B}_t = dW_t - \tanh(\tilde{X}_t) \, dt = d\tilde{X}_t - \tanh(\tilde{X}_t) \, dt. \tag{2}$$
By Girsanov's theorem, $(\tilde{B}_t)_{t \leq T}$ is a Brownian motion with respect to the probability measure $\mathbb{Q}_T$,
$$d\mathbb{Q}_T := Z_T \, d\mathbb{P}, \tag{3}$$
where $T>0$ is fixed and
$$Z_T := \exp \left( \int_0^T \tanh(-x_0+W_s) \, dW_s - \frac{1}{2} \int_0^T \tanh^2(-x_0+W_s) \, ds \right).$$
Hence, by $(2)$, $(\tilde{X}_t)_{t \leq T}$ is a weak solution to $$d\tilde{X}_t = \tanh(\tilde{X}_t) \, dt + d\tilde{B}_t, \qquad \tilde{X}_0 = -x_0$$ on the probability space $(\Omega,\mathcal{A},\mathbb{Q}_T)$. Since $\tanh(x)=-\tanh(-x)$ we find that $X_t := -\tilde{X}_t$  is a weak solution to $$dX_t = - \tanh(X_t) \, dt + dB_t, \qquad X_0 = x_0$$ with $B_t := - \tilde{B}_t$. 
Step 2: We need to compute
$$\mathbb{E}_{\mathbb{Q}_T}(X_T) = - \mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T).$$
By Itô's formula, we have
$$d(\log(\cosh(-x_0+W_t)) = \tanh(-x_0+W_t) \, dW_t + \frac{1}{2} (1-\tanh^2(-x_0+W_t)) \, dt,$$
and so
\begin{align*}&\int_0^T \tanh(-x_0+W_t) \, dW_t - \frac{1}{2} \int_0^T \tanh^2(-x_0+W_t) \, dt \\ &= \log(\cosh(-x_0+W_T)) - \log(\cosh(-x_0))- \frac{T}{2}.\end{align*}
This implies
$$Z_T = \frac{\cosh(-x_0+W_T)}{\cosh(x_0)} \exp \left(- \frac{T}{2} \right). \tag{4}$$
Consequently, we conclude that
$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) \stackrel{(3)}{=} \mathbb{E}(Z_T \tilde{X}_T) \stackrel{(4)}{=} \exp \left(- \frac{T}{2} \right) \frac{1}{\cosh(x_0)} \mathbb{E}((-x_0+W_T) \cosh(-x_0+W_T)). \tag{5}$$
Since $W_T$ is Gaussian with mean zero and variance $T$, we have
\begin{align*} \mathbb{E}(W_T \cosh(-x_0+W_T)) &=\frac{1}{\sqrt{2\pi T}} \int_{\mathbb{R}} x \cosh(x-x_0) \exp \left(- \frac{x^2}{2T} \right) \, dx \\ &= - \frac{T}{\sqrt{2\pi T}} \int_{\mathbb{R}} \cosh(x-x_0) \frac{d}{dx} \exp \left( - \frac{x^2}{2T} \right) \, dx\end{align*}
and so, by the integration by parts formula,
\begin{align*} \mathbb{E}(W_T \cosh(-x_0+W_T)) &=- \frac{T}{\sqrt{2\pi T}} \int_{\mathbb{R}} \sinh(x-x_0) \exp \left( - \frac{x^2}{2T} \right) \, dx \\ &= -T \mathbb{E}(\sinh(-x_0+W_T)). \end{align*}
Consequently, we obtain from $(5)$ that
$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) = \exp \left(- \frac{T}{2} \right) \frac{1}{\cosh(x_0)} \left[ -x_0 \mathbb{E}(\cosh(x_0+W_T)) - T \mathbb{E}\sinh(-x_0+W_T) \right].$$
Writing $$\cosh(x) = \frac{e^x+e^{-x}}{2} \quad \text{and} \quad \sinh(x) = \frac{e^x-e^{-x}}{2}$$
and using the fact that the exponential moments of Gaussian random variables are known, we conclude that
$$\mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) =  \frac{1}{\cosh(x_0)} \left[ -x_0 \cosh(x_0) + T \sinh(x_0) \right] = -x_0 + T \tanh(x_0).$$
Hence,
$$\mathbb{E}_{\mathbb{Q}_T}(X_T) = - \mathbb{E}_{\mathbb{Q}_T}(\tilde{X}_T) = x_0 - T \tanh(x_0).$$
