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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.

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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.

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Simply integrate $\int -1/2\cos(2x)dx=-1/4\sin(2x)$ But i think you only have to do integration by parts one time. – Kaladin Mar 8 '14 at 0:30

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 $\sin 2x$?

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I forgot to take the 3/2 out of the second integral... Silly mistake. Got it! Thanks! – AlecLeonK Mar 8 '14 at 0:30

Based on what you obtained, $$-\int\frac{-1}{2}cos(2x)dx=\int\frac{1}{2}cos(2x)dx=\int\frac{1}{4}cos(2x)d(2x)=\frac{1}{4}\sin(2x)+C$$

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