Application of Ito Isometry

Question

1) Show that \begin{equation} cov(\int_0^T(\mu_t+v_t)dB_t,\int_0^T(\mu_t-v_t)dB_t=\int_0^T(\mathbb{E}[\mu_t^2]-\mathbb{E}[v_t^2])dt \end{equation} The proof of this I believe is fairly straightforward. By the Ito isometry inner product rule, \begin{equation} \mathbb{E}[(\int_0^T X_t dB_t)(\int_0^T Y_t dB_t)]=\mathbb{E}[\int_0^T X_t Y_t dt] \end{equation} Thus, \begin{equation} \mathbb{E}[(\int_0^T(\mu_t+v_t)dB_t)(\int_0^T(\mu_t-v_t)dB_t)]=\\ \mathbb{E}[\int_0^T(\mu_t+v_t)(\mu_t-v_t)dt]=\\ \mathbb{E}[\int_0^T(\mu_t^2-v_t^2)dt]=\int_0^T(\mathbb{E}[\mu_t^2]-\mathbb{E}[v_t^2])dt \end{equation} 2) Using this result, calculate \begin{equation} cov(\int_0^T(B_te^t+\sqrt{t})dB_t,\int_0^T(B_te^t-\sqrt{t})dB_t) \end{equation} So, I can easily follow exactly the same steps as above, and get $B_t^2 e^{2t}-t$ for the product of $(B_t e^t+\sqrt{t})(B_t e^t-\sqrt{t})$. However, this part of the problem includes a hint to use the result that $ab=\frac{1}{2}(a+b)^2-\frac{1}{2}a^2-\frac{1}{2}b^2$. Of course, going through all of the math of this leads to the exact same result as what I obtained above by doing an elementary FOIL in my head, but I feel like the hint must mean that the logic that needs to be used here to solve the problem must follow some different path than that. Any guidance would be much appreciated, unless it really is that simple.