CUET Preparation Today
Find the integral: $\int \frac{1 - \sin x}{\cos^2 x} \, dx$ |
$\tan x + \sec x + C$ $\sec x - \tan x + C$ $\tan x - \sec x + C$ $\cot x + \text{cosec } x + C$ |
$\tan x - \sec x + C$ |
The correct answer is Option (3) → $\tan x - \sec x + C$ We have $\int \frac{1 - \sin x}{\cos^2 x} \, dx = \int \frac{1}{\cos^2 x} \, dx - \int \frac{\sin x}{\cos^2 x} \, dx$ $= \int \sec^2 x \, dx - \int \tan x \sec x \, dx$ $= \tan x - \sec x + C$ |
