$\displaystyle F=(x^{2}+y^{3})i+(2xy)j$ $\displaystyle C:x^{2}+y^{2}=9$

I'm not totally sure how to do this problem. So far, I've done this:

$\displaystyle r(t)=rcos(t)i+rsin(t)j$ => $\displaystyle 3cos(t)i+3sin(t)j$ $\displaystyle 0\leq t \leq 2\pi$

$\displaystyle F=(9cos^{2}(t)+27sin^{3}(t))i+18sin(t)cos(t)j$

$\displaystyle \frac{dr}{dt}=-3sin(t)i+3cos(t)j$

$\displaystyle F * \frac{dr}{dt}=$=$\displaystyle -3sin(t)(9cos^{2}(t)+27sin^{3}(t))+3cos(t)(18sin(t) cos(t))$

=> $\displaystyle -27sin(t)cos^{2}(t)+81sin^{4}+48sin(t)cos^{2}(t)$

$\displaystyle \displaystyle \int^{2\pi}_{0}81sin^{4}+21sin(t)cos^{2}(t)dt$

From here, I don't know how to calculate the integral. Is there a trig identity that simplifies the problem? Do I have to integral by parts or substitution? I plugged it into wolframalpa and got a solution but it didn't give me what I really wanted, the steps on how to solve the integral.

Am I even approaching the problem correctly? Thanks