I am doing Oksendal's book exercises one by one. I got stuck in 3.2.
I need to prove, from the definition that
$$\int_{0}^{t}B_s^2\text{d}B_s=\frac{B_s^3}{3}-\int_{0}^{t}B_s\text{d}s,$$
where $B_s$ is Brownian motion and the integral is Ito's.
Denote $B_{t_i}$ by $B_i$, for a partition $\{t_i\}_{i=2}^n$.
$$B_t^3=\sum_{0}^{n-1}\left(B_{i+1}^3-B_{i}^3\right)=\sum_{0}^{n-1}\left(B_{i+1}^2+B_{i+1}B_i+B_i^2\right)\left(B_{i+1}-B_i\right)$$
From the third terms we get $\int_{0}^{t}B_s^2\text{d}B_s$. I could add and subtract $B_i$ to the $B_{i+1}$ to make appear sums like $\sum(B_{i+1}-B_i)^2$, which tend to the quadratic covariation ($=t$), but there remain still some sums, like $\sum (B_{i+1}-B_i)^3$ that I don't know how to handle.
Nevermind! I got the idea. I will keep the post so you can enjoy. I just looked again to the integral in the right-hand side and the $(t_{i+1}-t_i)$ that appear in the sums. Replaced them conveniently.