- ?? Online homework says this is wrong.
- ?? Online homework says this is also wrong.
I agree with your choices of u and dv (though I would put $\displaystyle \begin{align*} \mathrm{d}\theta \end{align*}$ at the end of the $\displaystyle \begin{align*} \mathrm{d}u \end{align*}$ term.
So you need to remember that integration by parts is as follows: $\displaystyle \begin{align*} \int{ u\,\mathrm{d}v} = u\,v - \int{ v\,\mathrm{d}u} \end{align*}$, so that means with your choices of $\displaystyle \begin{align*} u \end{align*}$ and $\displaystyle \begin{align*} \mathrm{d}v \end{align*}$ you should get $\displaystyle \begin{align*} \int{ 3\theta\cos{(\theta)}\,\mathrm{d}\theta} = 3\theta\sin{(\theta)} - \int{ 3\sin{(\theta)}\,\mathrm{d}\theta} \end{align*}$. Go from here.
NO! $\displaystyle \begin{align*} u\,v = 3\theta\sin{(\theta)} \end{align*}$ NOT $\displaystyle \begin{align*} 3\sin^2{(\theta)} \end{align*}$.
Surely you can integrate $\displaystyle \begin{align*} 3\sin{(\theta)} \end{align*}$...