Solve $\int\ 3x\cos(2x)dx$ with integration by parts I have this problem here that is asking for me to use Integration by Parts.
I solved it out but it seems as if it can be an ongoing function. I was told you can work with it if the second time you integrate by parts it will be similar to the original integral but I don't see how to get it.

My answer after doing integration by parts twice was:
$$\int\ 3x\cos(2x)dx=\frac{3x}{2}\sin(2x)+\frac{3}{4}\cos(2x)-\int\frac{-1}{2}\cos(2x)dx$$
Any help on figuring out what to do from here would be very grateful thank you.

Only one application is necessary; letting $u = 3x$ and $dv = \cos 2x dx$, so that $du = 3 dx$ and $v = \frac 1 2 \sin 2x$, we find that
$$\int 3x \cos(2x) dx = \frac{3x}{2} \sin 2x - \frac 3 2 \int \sin 2x dx$$
Now can you find the antiderivative of $\sin 2x$?
