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Without throwing too much technical details, The Ito isometry says

$$ \mathbb{E}\bigg[\Big(\int^t_0 X_s d W_s\Big) \Big(\int^t_0 X_s d W_s\Big)^T \bigg] = \int^t_0 X_s\,X_s^T ds, $$

for some suitable Brownian motion $W_t$.

How about in the above two integrals $t$ are not the same? That is,

$$ \mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_0 X_s d W_s\Big)^T \bigg] = ? $$

Is it equal to $\int^{\mathrm{min}(t_1, t_2)}_0 X_s\,X_s^T ds$? If yes how to prove?

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Well your intuition is right here is why, supposing $t_1<t_2$ :

$$\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_0 X_s d W_s\Big)^T \bigg] =\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_1}_0 X_s d W_s+\int^{t_2}_{t_1} X_s d W_s\Big)^T \bigg]=$$ $$\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_1}_0 X_s d W_s\Big)^T \bigg]+\mathbb{E}\bigg[\Big(\int^{t_1}_0 X_s d W_s\Big) \Big(\int^{t_2}_{t_1} X_s d W_s\Big)^T \bigg]$$

Now observe that the second term is null as the two random variables are independent as the first integral belongs to $\mathcal{F}_{t_1}$ and the second to $\mathcal{F}_{t_2, t_1}$ that are independent. You are left with the term for which you know the result already.

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