what is the answer of this stochastic integral?

Question

as we know "ito integral "$$\int_{0}^{t}B_sdB_s=\frac{1}{2}B_s^2-\frac{1}{2}t$$now, I am searching for the solution for this one :$$\int_{0}^{t}B_s^2dB_s$$or$$\int_{0}^{t}B_s^4dB_s$$ $B_t$ is standard Brownian motion

Answer 1

Below I assuming that $\displaystyle B_s$ is a standard Brownian motion i.e. a Wiener process. Let's start with $\displaystyle \int_0^tB^2_sdB_s$. Choose $\displaystyle g(B_s)=B_s^3$, by Ito's lemma we have $$ dg(B_s)=3B_s^2dB_s+3B_sds $$ thus taking integrals from both sides we get $$ \int_0^tdg(B_s)=3\int_0^tB_s^2dB_s+3\int_0^tB_sds $$ or $$ \int_0^tB_s^2dB_s=\frac13B_t^3-\int_0^tB_sds $$ For $\displaystyle \int_0^tB^4_sdB_s$, define $\displaystyle g(B_s)=B_s^5$ for which by Ito's lemma we have $$dg(B_s)=5B_s^4dB_s+10B^3_sds$$ therefore $$B_t^5=5\int_0^tB_s^4dB_s+10\int_0^tB^3_sds$$