Why is my answer for integrating tan^3(x) dx wrong?

[LusTRouZ]

I'm not understanding why my answer is wrong

Homework Statement

$$\int$$tan^3(x) dx
This is the solution I've been getting for this problem, but I notice you get a different answer when you let u = tan(x) and du = sec^2(x) dx

Homework Equations

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

The Attempt at a Solution

$$\int$$tan(x) (sec^2(x) - 1) dx
$$\int$$tan(x) sec^2(x) - $$\int$$tan(x) dx
. // u = sec(x) du = sec(x)tan(x) dx
$$\int$$u du - $$\int$$sin(x) dx$$/$$cos(x)
. // z = cos(x) zu = -sin(x) dx
$$\int$$u du - $$\int$$-zu$$/$$z
(1$$/$$2)u^2 - (-ln |z|)
(1$$/$$2)sec^2(x) + ln |cos(x)|

Homework Statement

Homework Equations

The Attempt at a Solution

 

[Char. Limit]

What you have there is correct, up to a constant at least.