Can this be simplified?
$$\frac{\int_a^b \int_c^d \frac{1}{l_1(y,z)l_2^2(y,z)} \cos(kl_1+kl_2) \,dy\,dz }{\int_a^b \int_c^d \frac{1}{l_1(y,z)l_2^2(y,z)} \sin(kl_1+kl_2) \,dy\,dz}$$
where $$l_1 =\sqrt{A^2+y^2+z^2}$$ $$l_2=\sqrt{B^2+(C-y)^2+z^2} $$
$A,B,C$ and $k$ are constants.
More Information:
Let $K_1=\int_a^b \int_{c(w)}^{d(w)} \frac{1}{l_1(y,z)l_2^2(y,z)} \cos(kl_1+kl_2) \,dy\,dz\,\,\,$ and $K_2=\int_a^b \int_{c(w)}^{d(w)} \frac{1}{l_1(y,z)l_2^2(y,z)} \sin(kl_1+kl_2) \,dy\,dz\,\,\,$
(I set $d(w) =w/2$ and $c(w)=-w/2$)
I wanted to maximize $K_1^2 + K_2^2$ with respect to $w$ so:
$$\frac{\partial}{\partial w}(K_1^2 + K_2^2)=2 K_1 \frac{\partial K_1}{\partial w}+2 K_2 \frac{\partial K_2}{\partial w}=0$$
This equation will allow me to find the extremum.
$$2 K_1 \frac{\partial K_1}{\partial w}+2 K_2 \frac{\partial K_2}{\partial w}=0$$
use Leibniz Integral Rule for $\frac{\partial K_1}{\partial w}$ and $\frac{\partial K_2}{\partial w}$
Dividing by $K_2$
$$\frac{K_1}{K_2} \frac{\partial K_1}{\partial w}+ \frac{\partial K_2}{\partial w}=0$$
$K_1$ and $K_2$ are very similar so I thought that there might be a mathematical trick to simplify $\frac{K_1}{K_2}$
At first, I thought I can get away with setting $ \frac{\partial K_1}{\partial w}=0$ and $\frac{\partial K_2}{\partial w}=0$ but it seems that there is a conflict in the two equations.
So I tried not to ignore $\frac{K_1}{K_2}$. One of the possible solutions is
$$\frac{K_1}{K_2}=-\frac{\sin(kl_1+kl_2)}{\cos(kl_1+kl_2)}$$