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.