Find the following integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$

(A) Given integral: $\int \frac{2-3 \sin x}{\cos ^{2} x} d x$
Split the fraction into two parts:
$= \int \left( \frac{2}{\cos ^{2} x} - \frac{3 \sin x}{\cos ^{2} x} \right) d x$
Using trigonometric identities $\frac{1}{\cos ^{2} x} = \sec ^{2} x$ and $\frac{\sin x}{\cos ^{2} x} = \tan x \sec x$:
$= \int 2 \sec ^{2} x \, d x - 3 \int \tan x \sec x \, d x$
Integrating the terms:
$= 2 \tan x - 3 \sec x + C$
where $C$ is an arbitrary constant.