Let $W = (Wt)_{t\geq0}$ be a standard one dimensional Brownian motion. Prove that $$\int_{0}^{t} W_s^2 dW_s= \frac{1} {3} W_t^3-\int_{0}^{t}W_sds$$ $$\int_{0}^{t} sdW_s=tW_t-\int_{0}^{t}W_sds$$ I found this exercise online while preparing for exam and I am clueless.The only thing I have noticed is that the term $ \int_{0}^{t}W_sds$ looks like $\int_{0}^{t}F'(X_s)dX_s$ from Ito theorem.
Could you help me with solution?
For the second one use integration by parts formula. The integration by parts formula states that if $X$ and $Y$ are two processes then
$$ d(X_t Y_t) = X_t dY_t+ Y_t dX_t + dX_t\cdot dY_t$$
In your case, we set $X=W_t$ and $Y_t=t$. Then
$$ d(W_t t) = W_t dt + t dW_t + dt\cdot dW_t$$
From "multiplication table," we know that $dt\cdot dW_t=0$. So that
$$ d(W_t t) = W_t dt + t dW_t$$
or, in the integral form, $$ t W_t = \int_0^tW_s ds+\int_0^t s dW_s $$