$\int \frac{(5 \sin \theta-2) \cos \theta}{\left(5-\cos ^2 \theta-4 \sin \theta\right)} d \theta=$

Substitute $t = \sin \theta$, so $dt = \cos \theta d\theta$. Using $\cos^2 \theta = 1 - t^2$, the integral becomes

$I = \int \frac{5t - 2}{t^2 - 4t + 4} dt$.

The denominator simplifies as $(t-2)^2$. Let $u = t - 2$, so $t = u + 2$ and $dt = du$. The integral becomes

$I = \int \frac{5u + 8}{u^2} du = 5 \ln|u| - \frac{8}{u} + C$.

Substitute back $u = t - 2$ and $t = \sin \theta$ to obtain

$I = 5 \ln|\sin \theta - 2| - \frac{8}{\sin \theta - 2} + C$.

Since $\sin \theta - 2 < 0$, it is also valid to write the result as

$I = 5 \ln(2 - \sin \theta) - \frac{8}{\sin \theta - 2} + C$.

Final answer: $\boxed{B}$