Finding the distribution of an Ito integral. $\int_0^t sB_s \, \mathrm{d}s$ I'm a little baffled by this, I'm supposed to find the distribution of $X_t$ where, $X_t=\int_0^t sB_s \,   \mathrm{d}s$. What I can think of is to consider the process
$$\begin{align}
Y_s &= s^2B_s \\\\
dY_s &= 2sB_sds+s^2dBs+sds \\\\
\int_0^t \, \mathrm{d}Y_s &= 2\int_0^t sB_s\, \mathrm{d}s+\int_0^t s^2\mathrm{d}Bs+\int_0^t s\, \mathrm{d}s \\\\
\frac{t^2 B_t-\int_0^t s^2dBs-\int_0^t sds }{2} &= \int_0^t sB_s\, \mathrm{d}s 
\end{align}$$
Now I know that the first element in the numerator is normal and the second one is normal too with the third being deterministic. Therefore I know that the whole thing is normal and I need to find the expectations and the variances. But I'm not sure how to go about finding the variance since they are correlated.
I appreciate any help.

Let $X_t = \int_0^t s B_s \mathrm{d}s$. The process $X_t$ is Gaussian as a linear functional of the Wiener process, which is Gaussian itself. Thus, the distribution of $X_t$ is determined by its two moments:
$$
    \mathbb{E}\left(X_t\right) \stackrel{\text{linearity}}{=} \int_0^t s \, \underbrace{\mathbb{E}\left(B_s\right)}_0 \, \mathrm{d}s  = 0.
$$
The variance of $X_t$ is just the second moment:
$$
   \mathbb{Var}\left(X_t\right) = \mathbb{E}\left(X_t^2\right) = \mathbb{E}\left( \int_0^t s_1 B_{s_1} \mathrm{d}s_1  \int_0^t s_2 B_{s_2} \mathrm{d}s_2 \right)  \stackrel{\text{linearity}}{=} \int_0^t \int_0^t s_1 s_2 \mathbb{E}\left(B_{s_1} B_{s_2}\right) \mathrm{d}s_1 \mathrm{d}s_2.
$$
Since $\mathbb{E}\left(B_{s_1} B_{s_2}\right) = \min(s_1, s_2)$, we have:
$$
  \mathbb{Var}\left(X_t\right) = \int_0^t \int_0^t s_1 s_2 \min(s_1, s_2) \mathrm{d}s_1 \mathrm{d}s_2 = \frac{2 t^5}{15}.
$$
Thus, $X_t \sim \mathcal{N}\left(0, \frac{2 t^5}{15} \right)$.
