How would you do this

integral of tan^3(x)dx

I broke it down into integral of tan(x)(sec^2(x) - 1)dx

u = tan(x)

du = sec^2(x)

so i got integral of u(du) - integral of (u)

I am not so sure if this is correct so far, and I do not know where to go from here. Thanks for the help.