Question

intergal sec2xtan2x dx​

Answer 1

Step-by-step explanation:

We have,

$\int \sec(2x) \tan(2x) dx$

$let \: \: 2x = t \\ \implies \: dx = \frac{dt}{2}$

$= \frac{1}{2} \int \sec(t) \tan(t) dt$

$= \frac{1}{2} \sec(t) + c$

$= \frac{1}{2} \sec(2x) + c$