I evaluated int tan^2(3x)dx to be (1/3)tan(3x)-x+c but my online homework marked it as incorrect, what did I miss?

Question

anonymous · Answer

yeah I got that its not correct, so what did I do wrong? what is the correct method?

anonymous · Answer

that's what I did, the integral of sec^2 is tan, right? and then the 1/3 coefficient comes because there's a (3x) and the integral of 1 is just x... so where was I wrong?

loser66 · Answer

read it http://www.wolframalpha.com/input/?i=int+%28tan^2+%283x%29+%29dx

anonymous · Answer

I did plug my 3x back in, the wolfram page showed me exactly what I got for my answer...

loser66 · Answer

if it so, your net is wrong.

loser66 · Answer

Can you see my comments? I cannot see anything but this box

anonymous · Answer

oh yes I can see them. thanks for verifying my answer, I'll have to email my instructor