How do you integrate (sec2x)^2 ?

Question

Hi, does anyone know how to calculate this integral???... sec^2(2x)? i.e (sec2x)^2?

Answer 1

Aye

∫sec²x = tanx,

So ∫sec²(2x) =1/2 tan2x.

If you prefer, you can make an intermediate step by changing the variables: 2x=u, 2dx = du, dx = 1/2 du.

Hope this helps!

Answer 2

hey buddy! do you know that the integration is just the reverse process of differentiation?

well, as known from differential the derivative of tan u = (sec u)^2 du therefore the integral of (sec u)^2 du= tan u + c

since the angle of your problem is 2x, you let u = 2x

so that taking du = 2 dx then dx = du/2

from the integral of (sec2x)^2 dx change the variable

it becomes integral of (sec u)^2 du/2 or (1/2)(sec u)^2 du

apply the formula from above

(1/2) tan u +c or (1/2) tan 2x + c

hope this helps you!

Answer 3

You can find this integral in a table of integrals:

Int (sec^2(x)) = tan x + C

You can then use the chain rule backwards to figure this out. Knowing that once we differentiate we're going to get an extra 2 (from the derivative of 2x). We know we have to pull that out of the integral.

The answer would then be:
(tan 2x)/2 + C

Answer 4

Let's let u = 2x, x = u/2, dx = du/2.
Then we have
1/2 * ∫ sec² u du = 1/2 tan u = 1/2 tan 2x +C.

Answer 5

we know that the derivitive of tan (2x) +C is = (sec(2x))^2 times two cos of the chain rule

so to get rid of the two we put a half in front of the tan so the derivitive becomes (sec(2x))^2

IE: d/dx 1/2 tan(2x) +C
=1/2 * (sec(2x))^2 * 2

Answer 6

When I ask the other two I work with for their best, they answer, 'Huh?' This is way down south, my friend. The most I can hope for is an intelligent nod of someone's head---no speech.