Consider the Abelian Higgs model in $\mathbb{R}^2$ with the action, $$E= \frac{1}{2}\int_{\mathbb{R}^2}F_{A}^2+|D_A \phi|^2 + \frac{1}{4}(1-|\phi|^2)^2 dx$$ where $A=A_1dx_1+A_2dx_2$ is a 1-form connection, $\phi=\phi_1+i\phi_2$ is a complex-valued scalar field. $F_A$ is defined to be $\partial_1A_2-\partial_2A_1$ and $D_A=d\phi-iA\phi$.
Bogomolny pointed out that the action can be rewritten as $$E= \frac{1}{2}\int_{\mathbb{R}^2}|(\partial_1 \phi_1+A_1\phi_2)-(\partial_2\phi_2+A_2\phi_1)|^2 +|(\partial_2 \phi_1+A_2\phi_2)+(\partial_1\phi_2-A_1\phi_1)|^2 \\+|F_{A}+\frac{1}{2}(|\phi|^2-1)|^2+F_{A}dx$$
I am trying to do so but failed.
Expressing the term $|D_A \phi|^2$, $$|D_A \phi|^2=|\partial_i\phi-iA_i\phi|^2=|\partial_i\phi_1+i\partial_i\phi_2-iA_i\phi_1+A_i\phi_2|^2=|\partial_i\phi_1+A_i\phi_2|^2+|\partial_i\phi_2-A_i\phi_1|^2$$ Next I consider $$F_{A}^2+\frac{1}{4}(1-|\phi|^2)^2=(F_{A}-\frac{1}{2}(1-|\phi|^2))^2+F_{A}-F_{A}|\phi|^2$$ Thus, by combining the term \begin{align*} F_{A}^2+|D_A \phi|^2 + \frac{1}{4}(1-|\phi|^2)^2 &=(F_{A}-\frac{1}{2}(1-|\phi|^2))^2+F_{A}-F_{A}|\phi|^2\\ &+|\partial_1\phi_1+A_1\phi_2|^2+|\partial_1\phi_2-A_1\phi_1|^2\\ &+|\partial_2\phi_1+A_2\phi_2|^2+|\partial_2\phi_2-A_2\phi_1|^2\\ &=(F_{A}-\frac{1}{2}(1-|\phi|^2))^2+F_{A}-F_{A}|\phi|^2\\ &+((\partial_1\phi_1+A_1\phi_2)-(\partial_2\phi_2-A_2\phi_1))^2\\ &+((\partial_2\phi_1+A_2\phi_2)+(\partial_1\phi_2-A_1\phi_1))^2\\ &+2(\partial_1\phi_1+A_1\phi_2)(\partial_2\phi_2-A_2\phi_1)\\ &-2(\partial_2\phi_1+A_2\phi_2)(\partial_1\phi_2-A_1\phi_1) \end{align*} Therefore, my job is to show that $$\int_{\mathbb{R}^2}2(\partial_1\phi_1+A_1\phi_2)(\partial_2\phi_2-A_2\phi_1)-2(\partial_2\phi_1+A_2\phi_2)(\partial_1\phi_2-A_1\phi_1)-F_{A}|\phi|^2=0$$ I can only simplify the above term to $$\int_{\mathbb{R}^2}\partial_1(|\phi|^2A_2)-\partial_2(|\phi|^2A_1)+2(\partial_2\phi_1\partial_1\phi_2-\partial_1\phi_1\partial_2\phi_2)$$ How can I do this?