What is the integral... ∫ [1+tan^2(2x)] dx

what is ∫[1+tan^2 (2x)] dx?

i know that 1+tan^2=sec^2 but i dont know if that is on the right track
i tried using "u substitution" too and received an answer of ((sec(2x))^3)/6. which i dont think is right

Thank you for helping!!!

1 + tan^2(2x) = sec^2(2x)

It doesn't matter what the value is inside the parenthesis; it stil follows the same trig identity.
And we also know that d/dx(tanA) = sec^2(A) right? so....

∫[1 + tan^2(2x)]dx
= ∫sec^2(2x)dx

u = 2x
du = 2dx
dx = (1/2)du

= (1/2) ∫sec^2(u)du
= (1/2) tan(u) + C

plug 2x back in for u.
= (1/2)tan(2x) + C

Hope this helps :D

∫ [1+tan^2(2x)] dx
= ∫ [ sec^2(2x)] dx
= (1/2) tan (2x) + C

= (1/2)tan(2x)+C

1/2 Tan[2 x]