what is the anti derivative of: sec^2(x)*tan^2(x)?

Answer 1

∫sec^2 x tan^2 x dx, let u = tan x, du = sec^2 x dx
= ∫u^2 du
= u^3 / 3 + c
= (1/3) tan^3 x + c.

Answer 2

Notice that
(tan x)' = sec²x.
Substitute t = tan x. Then
∫sec²x tan²x dx
= ∫ t² dt
= t³/3 + C
= tan³x/3 + C

Answer 3

hello as you know sec^2(x) = 1 / cos^2(x) = 1 + tan^2(x) so the function is equal to (1 + tan^2(x))(tan^2(x)) so you have derivative of the tan(x) in your function sth like f`(x)*f(x) so the integral of this function is 1/3 tan^3(x) + C