Find the integral: $\int \frac{1 - \sin x}{\cos^2 x} \, dx$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $\tan x - \sec x + C$
We have
$\int \frac{1 - \sin x}{\cos^2 x} \, dx = \int \frac{1}{\cos^2 x} \, dx - \int \frac{\sin x}{\cos^2 x} \, dx$
$= \int \sec^2 x \, dx - \int \tan x \sec x \, dx$
$= \tan x - \sec x + C$