TITLE: How to calculate $E[(\int_0^t{W_sds})^n], n \geq 2$
QUESTION [5 upvotes]: Let $W_t$ be a standard one dimension Brownian Motion with $W_0=0$ and $X_t=\int_0^t{W_sds}$.
With the help of ito formula, we could get
$$E[(X_t)^2]=\frac{1}{3}t^3$$ 
$$E[(X_t)^3]=0$$
When I try to employ the same method to calculate the general case $E[(X_t)^n]$, I got stuck.
I guess $X_t$ should be normal distribution since it could be the limit of the following
$$\lim_{n\rightarrow \infty}{\sum_{i=0}^{n-1}{W_{t_i}(t_{i+1}-t_i)}},$$
where $ W_{t_i}\sim norm(0,\sqrt{\frac{t_i}{n}}).$
If it is true, the problem would be trivial.
Update: Thanks for all the suggestions. Now I believe $X_t$ is a Gaussian process.
How about for this integral
$$Y_t=\int_0^t{f(W_s)ds}$$
if we assume that $f$ is some good function, say polynomial or exponential, i.e 
$$Y_t=\int_0^t{e^{W_s}ds}$$
$$Y_t=\int_0^t{[a_n(W_s)^n+a_{n-1}(W_s)^{n-1}+...+a_0]ds}$$

REPLY [5 votes]: The random variable $X_t$ is Gaussian for the reasons in Didier's answer.
You can calculate the variance directly (without Ito's formula) as follows:
$$\mathbb{E}\left[\left( \int^t_0 W_s ds \right)^2\right] 
= \int^t_0 \int^t_0  \mathbb{E}(W_r W_s)  dr ds 
= \int^t_0 \int^t_0  (r\wedge s) dr ds ={t^3\over 3}.$$