\(\int\displaystyle\frac{dx}{1+\cos 2x}\) is equal to
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \(\displaystyle \frac{1}{2}\tan x+C\)
Using the double-angle identity \(1+\cos 2x = 2\cos^2x\):
\( \displaystyle \frac{1}{1+\cos 2x} = \frac{1}{2\cos^2x} = \frac{1}{2}\sec^2x \)
\( \displaystyle \int\frac{1}{2}\sec^2x\,dx = \frac{1}{2}\tan x+C \)
Why the other options are incorrect
Option (2) \(2\tan x+C\) — Incorrect. The factor 2 of \(2\cos^2x\) is in the denominator, so it gives \(\frac{1}{2}\), not 2.
Option (3) \(x+\frac{1}{2}\sin 2x+C\) — Incorrect. This is the integral of \(1+\cos 2x\) itself, not of its reciprocal.
Option (4) \(-\frac{1}{2}\cot x+C\) — Incorrect. This is the integral of \(\frac{1}{1-\cos 2x} = \frac{1}{2}\mathrm{cosec}^2x\).