Question

What is `int sec^2(5x) dx`?

Answer 1

`intsec^2(5x)dx = 1/5tan(5x)+C`

Explanation:

Noting that `d/dx tan(x) = sec^2(x)`

Let `u = 5x => du = 5dx => dx = 1/5du`

`intsec^2(5x)dx = intsec^2(u)/5du`

`=1/5intsec^2(u)du`

`=1/5tan(u)+C`

`=1/5tan(5x)+C`