Will Ito's Isometry result in $E\left(\int_0^t \cos(u)\,dB_u \int_0^t \sin(u)\, dB_u \right) = E\left(\int_0^t \cos(u) \sin(u)\, du \right)$?

Question

If I have two integrals, $X_t = \int_0^t \cos(u)\,dB_u$and $Y_t = \int_0^t \sin(u)\, dB_u$ , where $B_u$ is a Wiener Process and I am trying to find:

$$ E\left(\int_0^t \cos(u)\,dB_u \int_0^t \sin(u)\, dB_u \right) $$

I am wondering if I can apply Ito's Isometry to obtain the relation: $$ E\left(\int_0^t \cos(u)\,dB_u \int_0^t \sin(u)\, dB_u \right) = E\left(\int_0^t \cos(u) \sin(u)\, du \right) $$

I know that this relation is true but don't know how to get it. Thanks!

Answer 1

Hint:

Apply the polarization identity in $\mathbb{R}$: $$xy = \frac{1}{4}((x+y)^2-(x-y)^2),$$ to $x = \displaystyle \int_{0}^{t} \cos(u) dB_u$ and $y = \displaystyle \int_{0}^{t} \sin(u) dB_u$, together with the Itô-isometry. This will lead to the result.