∫ xsec^2 2x dx is equal to

The correct answer is Option (4) → \(\displaystyle\frac{x}{2}\tan 2x+\frac{1}{4}\log|\cos 2x|+C\)

Using integration by parts with x as the first function and \(\sec^2 2x\) as the second (\(\int\sec^2 2x\,dx = \frac{1}{2}\tan 2x\)):

    \( \displaystyle \int x\sec^2 2x\,dx = \frac{x}{2}\tan 2x-\frac{1}{2}\int\tan 2x\,dx \)

Since \(\int\tan 2x\,dx = -\frac{1}{2}\log|\cos 2x|\):

    \( \displaystyle \int x\sec^2 2x\,dx = \frac{x}{2}\tan 2x+\frac{1}{4}\log|\cos 2x|+C \)

Why the other options are incorrect

• Option (1) \(\displaystyle\frac{x}{2}\tan 2x+\frac{1}{4}\log|\sec 2x|+C\) — Incorrect. \(\log|\sec 2x| = -\log|\cos 2x|\), so the sign of the log term is wrong.

• Option (2) \(\displaystyle\frac{x^2}{2}\tan 2x+C\) — Incorrect. A product cannot be integrated factor by factor; its derivative is \(x\tan 2x+x^2\sec^2 2x\).

• Option (3) \(\displaystyle 2x\tan 2x+4\log|\cos 2x|+C\) — Incorrect. It multiplies by 2 where it should divide by 2 at each integration.