Prove the following equality without Ito's formula:
$$tB_t=\int_0^t sdB_s+\int_0^tB_sds.$$
I was thinking of moving $\int_0^t sdB_s$ to the left side and then taking the expectation on both sides to use the Itô isometry, but I don't know if that would go anywhere.
Let $0 = t_0 < t_1 < \ldots < t_n = t$ be a partition of $[0, t]$. Then
\begin{align*} t B_t &= \sum_{i=1}^{n} (t_i B_{t_i} - t_{i-1} B_{t_{i-1}}) \\ &= \sum_{i=1}^{n} t_{i-1}(B_{t_i} - B_{t_{i-1}}) + \sum_{i=1}^{n} B_{t_i}(t_i - t_{i-1}). \end{align*}
As the partition gets finer, the first sum converges in probability to $\int_{0}^{t} s \, \mathrm{d}B_s$ and the second term converges a.s. to $\int_{0}^{t} B_s \, \mathrm{d}s$, proving the desired equality.