\(\int\displaystyle\frac{dx}{1+\sin x}\) is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\tan x-\sec x+C\)
Multiplying numerator and denominator by \(1-\sin x\):
\( \displaystyle \frac{1}{1+\sin x} = \frac{1-\sin x}{1-\sin^2x} = \frac{1-\sin x}{\cos^2x} \)
\( \displaystyle = \sec^2x-\sec x\tan x \)
Using \(\int\sec^2x\,dx = \tan x\) and \(\int\sec x\tan x\,dx = \sec x\):
\( \displaystyle \int\frac{dx}{1+\sin x} = \tan x-\sec x+C \)
Why the other options are incorrect
Option (1) \(\sec x+\tan x+C\) — Incorrect. Its derivative is \(\sec x\tan x+\sec^2x\), which is \(\frac{1+\sin x}{\cos^2x}\), not \(\frac{1-\sin x}{\cos^2x}\).
Option (2) \(\sec x-\tan x+C\) — Incorrect. It is the negative of the answer; its derivative is \(-\frac{1}{1+\sin x}\).
Option (4) \(\tan\frac{x}{2}+C\) — Incorrect. Its derivative is \(\frac{1}{2}\sec^2\frac{x}{2} = \frac{1}{1+\cos x}\), the result for \(1+\cos x\), not \(1+\sin x\).