Expected value of SinhX A random variable X is distributed according to a normal distribution with mean u and variance d.How to evaluate the expected value of SinhX?Thanks
 A: If $X$ is normal $(\mu,\sigma^2)$ then $X=\mu+\sigma Z$ where $Z$ is standard normal hence, by definition of $\sinh$, 
$$
E(\sinh X)=\tfrac12(\mathrm e^\mu E(\mathrm e^{\sigma Z})-\mathrm e^{-\mu}E(\mathrm e^{-\sigma Z})).
$$ 
The distribution of $Z$ is symmetric hence $E(\mathrm e^{\sigma Z})=E(\mathrm e^{-\sigma Z})$, which leaves us with 
$$
E(\sinh X)=E(\mathrm e^{\sigma Z})\sinh\mu.
$$ 
Now, the density $\varphi$ of $Z$ is defined by $\varphi(z)=(2\pi)^{-1/2}\mathrm e^{-z^2/2}$ and 
$$
\mathrm e^{\sigma z}\varphi(z)=\varphi(z-\sigma)\mathrm e^{\sigma^2/2}.
$$
(This algebraic identity is the (only) crucial step in the proof, can you check it?) 
Thus,
$$
E(\mathrm e^{\sigma Z})=\int_\mathbb R\varphi(z-\sigma)\mathrm e^{\sigma^2/2}\mathrm dz\stackrel{(z=u+\sigma)}{=}\mathrm e^{\sigma^2/2}\int_\mathbb R\varphi(u)\mathrm du=\mathrm e^{\sigma^2/2},
$$
and finally,
$$
E(\sinh X)=\mathrm e^{\sigma^2/2}\sinh\mu.
$$
A: If your random variable $X \sim N(0,1)$, then trivially $E[sinh(X)] = 0$, since $sinh(.)$ is symmetrical about zero:

(source: tri.org.au)
For the $X \sim N(\mu, \sigma^2)$ case that you seek, how would I do it? Quickest approach is to pop it into a computer algebra system. For your example, $X$ has pdf $f(x)$:

(source: tri.org.au)
and you seek:

(source: tri.org.au)
where I am using the Expect function from the mathStatica package for Mathematica. [I should perhaps add that I am one of the authors of the former.] All done.
If you wanted to do it manually, one could note that $sinh(X) = \frac12 (e^X - e^{-X})$, and $E[e^X]$ is just the mean of a Lognormal random variable, i.e.  $E[e^X] = exp({\mu +\frac{\sigma ^2}{2}})$, and similarly  $E[e^{-X}] = exp({-\mu +\frac{\sigma ^2}{2}})$, ... which will get you to the same place.
