Let $W$ be a Brownian motion defined on an appropriate probability space. I wish to determine the value of $$\int_0^t W_s^2\,dW_s$$ A partial solution follows: Let $f(x) = x^3/3$ and then apply Ito's Chain rule. Then, $$df(W_t) = f(W_0) + W_t^2\,dW_t + W_t\,dt$$ $$W_t^3/3 = \int_0^tW_s^2\,dW_s + \int_0^t W_s\,ds$$ $$W_t^3/3 - \int_0^t W_s\,ds = \int_0^tW_s^2\,dW_s $$
My question is the following: Can any further simplification be carried out from here? I tried the following: Recall the integration by parts formula (for continuous processes) $$X_tY_t = X_0Y_0 + \int_0^t X_s\,dY_s + \int_0^t Y_s\,dX_s + \langle X,Y \rangle_t $$ and now put $X = W$, and $Y_t = t$. Then, $$tW_t = \int_0^t W_s\,ds + \int_0^t s\,dW_s + \langle W_t,t \rangle_t$$