I am trying to express the following summation as integrals where $x \in [0,180^\circ]$,$x_i \in [0,180^\circ]$,$y \in [0,360^\circ]$ and $y_i \in [0,360^\circ]$. Will appreciate any help. Thank you. $$ f_1(x,y) = \sum_{i=1}^{n} \left[ \vphantom{\frac{a}{b}} \cos (x) \sin \left(x_i\right) \cos (y-y_i) - \cos\left(x_i\right) \sin(x) \right]^2, $$ $$ f_2(x,y) = \left\{\sum_{i=1}^{n} \vphantom{\frac{a}{b}} \sin\left(x_i\right) \sin (y-y_i) \left( \vphantom{\frac{a}{b}} \cos (x) \sin \left(x_i\right) \cos (y-y_i) - \cos\left(\theta_\ell\right) \sin(\theta) \right) \vphantom{\frac{a}{b}} \right\}^2, $$
$$ f_3(x,y) = \sum_{i=1}^{n} \vphantom{\frac{a}{b}} \sin^2\left(x_i\right) \sin^2 (y-y_i) . $$
Expand the arguments of the form $y-y_i$ by means of the addition formulas and develop the squares. This will allow you to take out factors $\text{sin/cos}(x/y)$, times summations of products $\text{sin/cos}(x_i/y_i)$.
For example (using an abbreviated notation),
$$f_3=\sum s_i^2(sc'_i-cs'_i)^2=\sum s_i^2(s^2c_i'^2-2sc_i'cs_i'+c^2s_i'^2)\\ =s^2\sum s_i^2c_i'^2-2sc\sum s_i^2s_i'c_i+c^2\sum s_i^2s_i'^2.$$
Unless the $x_i$ and $y_i$ form arithmetic progressions, you can't simplify this further.