Question

question 16
evaluate: $\int\sec 2x\tan 2xdx$.
$\frac{1}{2}\tan(2x)+c$
$-\frac{1}{2}\tan(2x)+c$
$\frac{1}{2}\sec(2x)+c$
$-2\sec(2x)+c$
no correct answer choice is given.

Explanation:

Step1: Use substitution

Let \(u = 2x\), then \(du=2dx\), and \(dx=\frac{1}{2}du\).
The integral \(\int\sec(2x)\tan(2x)dx\) becomes \(\frac{1}{2}\int\sec(u)\tan(u)du\).

Step2: Recall the integral formula

We know that \(\int\sec(u)\tan(u)du=\sec(u)+C\).
Substituting back \(u = 2x\), we get \(\frac{1}{2}\sec(2x)+C\).

Answer:

\(\frac{1}{2}\sec(2x)+C\) (corresponding to the third option)