∫(1-sin x)/(cos^2x) dx is equal to:(where c is an arbitrary constant)

The correct answer is Option (4) → \( \tan x-\sec x+c \)

The fraction splits into two standard integrands:

    \( \displaystyle \frac{1-\sin x}{\cos^2x}=\frac{1}{\cos^2x}-\frac{\sin x}{\cos x}\cdot\frac{1}{\cos x} \)

    \( \displaystyle =\sec^2x-\sec x\tan x \)

Using \( \int\sec^2x\,dx=\tan x \) and \( \int\sec x\tan x\,dx=\sec x \):

    \( \displaystyle \int(\sec^2x-\sec x\tan x)\,dx=\tan x-\sec x+c \)

Why the other options are incorrect

• Option (1) \( \cot x-\mathrm{cosec}\,x+c \) — Incorrect. \( \cot x \) and \( \mathrm{cosec}\,x \) arise when the denominator is \( \sin^2x \), not \( \cos^2x \).

• Option (2) \( \mathrm{cosec}\,x-\cot x+c \) — Incorrect. The denominator is \( \cos^2x \), so cosec and cot do not appear.

• Option (3) \( \sec x-\tan x+c \) — Incorrect. This has both signs reversed: \( \frac{d}{dx}(\sec x-\tan x)=\sec x\tan x-\sec^2x \), the negative of the integrand.