The value of integral I=∫(1-sinx)/(cos^2x)dx is :

The correct answer is Option (1) → $\tan x - \sec x + c$, where c is a constant

$I=∫\frac{1-\sin x}{\cos^2x}dx$

$=∫\frac{1}{\cos^2x}-\frac{\sin x}{\cos x}×\frac{1}{\cos x}dx$

$=∫\sec^2x-\sec x\tan xdx$

$=\tan x-\sec x+c$