integral of Sec[2x]Tan[2x]

i know u is sec 2x
du=2sec2xtan2x dx
what would i have to multiply with du so it would equal tan 2x dx?

if my question is confusing, then here's another example of what i'm talking about:
integral of (3x-2)^30 dx
u=3x-2
du=3 dx
1/3 du=dx (i need help with this step on the problem obove)
then i would substitute and solve the integral.

Well, dx= du/(2sec2x tan 2x) from your equation.

INT 2sec2x tan2x * dx= INT du
INT du= u

On the second.
You have u^30 du/3 since dx=du/3
INT u^30 du/3 = 1/3 INT u^30 du=
1/3 * 31 u^31

then on both, substitute back in the f(x) for u.

To find what to multiply with du so that it equals tan 2x dx, we need to manipulate the given expression:

du = 2sec^2(2x)tan(2x) dx

First, let's rewrite sec^2(2x) as 1/cos^2(2x) since tan(2x) = sin(2x)/cos(2x):

du = 2(1/cos^2(2x)) (sin(2x)/cos(2x)) dx

Now, let's simplify this expression by multiplying through by cos^2(2x):

du = 2sin(2x) dx

Therefore, to have du equal to tan(2x) dx, you would need to multiply du by (1/2)sin(2x) in the given integral.

Regarding the second example,

1/3 du = dx

To derive this, remember that u = 3x - 2 and du = 3 dx. To isolate dx, divide both sides of du = 3 dx by 3:

du/3 = dx

Therefore, 1/3 du is equal to dx, which you can use as the substitution for dx in the given integral.