The value of the integral \(\int\displaystyle\frac{1-\sin x}{\cos^2 x}\,dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → tan x − sec x + C, where C is an arbitrary constant.
Split the fraction into two standard integrands:
\( \displaystyle \frac{1-\sin x}{\cos^2 x} = \frac{1}{\cos^2 x} - \frac{\sin x}{\cos x}\cdot\frac{1}{\cos x} \)
\( = \sec^2 x - \sec x\tan x \)
Since \(\frac{d}{dx}(\tan x) = \sec^2 x\) and \(\frac{d}{dx}(\sec x) = \sec x\tan x\):
\( \displaystyle \int(\sec^2 x - \sec x\tan x)\,dx = \tan x - \sec x + C \)
Why the other options are incorrect
Option (1) sec x − tan x + C, where C is an arbitrary constant. — Incorrect. Both signs are reversed: its derivative is \(\sec x\tan x - \sec^2 x\), the negative of the integrand.
Option (3) sec x tan x + C, where C is an arbitrary constant. — Incorrect. Its derivative is \(\sec x\tan^2 x + \sec^3 x\), not the integrand.
Option (4) tan x + sec x + C, where C is an arbitrary constant. — Incorrect. Its derivative is \(\sec^2 x + \sec x\tan x = \frac{1+\sin x}{\cos^2 x}\), with + sin x instead of − sin x.