[Integral](Tan[x]/Cos[x]^2)dx

We've got [integral]f(u(x)) u'(x) dx -> [integral]f(u) du, u = tan(x)

So... how come the answer is not (Tan[x]^2)/2

the answer is (Tan[x])^2 /2 + C

∫tan(x)*sec²(x) dx = tan²(x)/2 + C

Your answer is correct.