My Stochastic calculus professor always used to say "When in doubt use Ito"

So let $f(t,x) = t x $ and compute $\partial_t f(t,x) = x$, $\partial_x f(t,x) = t $ and $ \partial_{xx} f(t,x) = 0$

Now the Ito lemma says for $f$ twice differentiable with respect to $x$ and once differentiable with respect to $t$ then the following formula holds for any Ito process:

$f(t, X_t) = f(0,X_0) + \int_{0}^{t}\partial_s f(s,X_s) ds + \int_{0}^{t}\partial_x f(s,X_s) dX_s$ $+ \int_{0}^{t}\partial_{xx} f(s,X_s) d \left< X,X \right>_s $

So applying the above fact to the function $f(t,x) = tx$ gives:

$t X_t = 0 + \int_{0}^{t}X_s ds + \int_{0}^{t} s dX_s + 0$ or to give you a starting answer
$\int_{0}^{t}X_s ds = t X_t -  \int_{0}^{t} s dX_s$

So in vague words we have that The (Riemann) integral of an ito process is equal to the difference of two Gaussian processes which should again be Gaussian .... (if all this logic is correct it should then suffice to characterize the processes covariance structure in order to have a complete understanding of the density of the process for instance)

For example one can compute the variance if $X_t$ is standard Brownian motion:

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^2 \mathbb{E}[(X_t)^2] -2t \mathbb{E}[X_t \int_{0}^{t} s dX_s] + \mathbb{E}[(\int_{0}^{t}sdX_s)^2] $

By the Ito isometry we have $\mathbb{E}[(\int_{0}^{t}sdX_s)^2] = \int_{0}^{t}s^2ds = t^{3}/3$.

To compute $\mathbb{E}[X_t \int_{0}^{t} s dX_s]$ notice first that the Ito integral of a deterministic function is always a Gaussian process.  I am pretty sure $X_t$ and $\int_{0}^{t}s dX_s$ are independent but the proof escapes me at the moment (any help would be greatly appreciated...) but assuming Independence we have   $\mathbb{E}[X_t \int_{0}^{t} s dX_s] = 0$.  If the processes are not independent the best I can say at the moment is that  $\mathbb{E}[X_t \int_{0}^{t} s dX_s]$ is a positive semi-definite function. Assuming the easier of the two cases we have finally:

$\mathbb{E}[(\int_{0}^{t}X_s ds)^2] = t^3 + \frac{t^{3}}{3} = \frac{4t^3}{3}$

Computing the covariance $\mathbb{E}[\int_{0}^{t}X_s ds \int_{0}^{u}X_s ds]$ involves dealing with term $\mathbb{E}[\int_{0}^{t}s dX_s \int_{0}^{u}s dX_s]$ which the only guess I have is this expression might also be independent.