I'm trying to prove the following equality
$$ \int_0^T W(t) dt = \int_0^T (T-t) dW(t) $$
where $W(t)$ is a standard brownian motion. I'm been trying to make use of the fact, that $dt = dW(t) dW(t)$ (informally?), such that
$$ \int_0^T W(t) dt = \int_0^T W(t) dW(t) dW(t) $$
in which case $W(t) dW(t) = (T-t)$ which makes no sense to me. I've also tried with integration by parts, but my background in measure theory is heuristic at best, so i have hard time imagining what i'm looking for.
Could anyone give me a hint in the right direction? I've also been wondering, if i could make use of itô's lemma in this case, but without any luck.
Hint: Since the stochastic integral is a linear mapping, your claim is equivalent to
$$\int_0^T W_t \, dt = \int_0^T T \, dW_t - \int_0^T t \, dW_t = T W_T - \int_0^T t \, dW_t.$$
To prove this identity apply Itô's formula for the (time-dependent) function
$$f(t,x) := t \cdot x.$$