I am trying to solve the given integral with Brownian motion $$ \int_{i=0}^T B_t\,d(B_t)^2 $$ I have solved so far like this, $$ \int_{i=0}^T B_t\,d(B_t)^2 = \sum_{0}^T B_t(B_{t+1} - B_t)^2 $$ We can re-write $B_t$ as, $$ B_t = \frac{1}{2}(B_{t+1} + B_t) - \frac{1}{2}(B_{t+1} - B_t) $$ Substituting, $$ \sum_{0}^T B_t(B_{t+1} - B_t)^2 = \sum_{0}^T (\frac{1}{2}(B_{t+1} + B_t) - \frac{1}{2}(B_{t+1} - B_t))(B_{t+1} - B_t)^2 $$ $$ = \frac{1}{2}\sum_{0}^T(B_{t+1} + B_t)(B_{t+1} - B_t)^2 - \frac{1}{2}\sum_{0}^T(B_{t+1} - B_t)(B_{t+1} - B_t)^2 $$ $$ = \frac{1}{2}\sum_{0}^T (B_{t+1} + B_t)(B_{t+1} - B_t)(B_{t+1} - B_t) - \frac{1}{2}\sum_{0}^T(B_{t+1} - B_t)^3 $$ $$ = \underbrace {\frac{1}{2}\sum_{0}^T (B_{t+1}^2 - B_t^2)(B_{t+1} - B_t)}_{\text{1st part}} - \underbrace{ \frac{1}{2}\int_{0}^T(dB)^3}_{\text{2nd part}} $$ Here is where the problem arises,
In the 2nd part,
we know that $dBt^2 = dt$, but what would be $dBt^3$
Does it go towards 0?
In the first part, we know that,
$\int{(B_{t+1}^2 - B_t^2)} = Bt^2$ but what happends with the extra term - $\int{(B_{t+1} - B_t)}$
Can anyone please help me out??
When you follow the hint by @Khosrotash you don't need $(dB_t)^3\,:$ From $$ d(B_t)^2=2B_t\,dB_t+dt\,,\quad\text{ and }\quad d(B_t)^3=3B_t^2\,dB_t+3B_t\,dt $$ your integral becomes \begin{align} &\textstyle\int_0^TB_t\,d(B_t)^2=\textstyle\int_0^T2B_t^2\,dB_t+\int_0^TB_t\,dt\\[2mm] &=\textstyle\frac{2}{3}B_T^3-\int_0^TB_t\,dt\,. \end{align} If we want we can write the last term as $$ \textstyle\int_0^TB_t\,dt=\int_0^T(T-t)\,dB_t\,. $$