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Hello I am currently stuck at integration by parts. And I was wondering why at the second $u$ substitution I should choose $u=\cos 3x$ instead of the $\sin 2x$. Is there a specific rule which applies?

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Best, Sebastian

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Why not try it for yourself and see: let $u=\sin 2x$ and $dv=\cos 3xdx$. Then $du=2\cos 2x dx$ and $v=\frac13\sin 3x$, so

$$\int\cos 3x\sin 2x dx=\frac13\sin 2x\sin 3x-\frac23\int\sin 3x\cos 2x dx\;,$$

and

$$\begin{align*} \int\sin 3x\cos 2x dx&=\frac12\sin 3x\sin 2x-\frac32\int\cos 3x\sin 2x dx\\ &=\frac12\sin 3x\sin 2x-\frac32\left(\frac13\sin 2x\sin 3x-\frac23\int\sin 3x\cos 2x dx\right)\\ &=\int\sin 3x\cos 2x dx\;, \end{align*}$$

which is obviously true but not very helpful!

The general rule in this situation is that you should let the second $u$ be the part that came from the original $du$; otherwise you simply undo the first integration by parts, just as I did here. (Note that while you might not come up with the general rule on your own, you can discover something about the problem just by trying. Never be afraid to try something; first, it might work, and secondly, even if it doesn’t, you’ll have learned something.)

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  • $\begingroup$ Yes that is probably the best way to go about those integrations ;) Thanks for your help, I will review my work :) $\endgroup$ – djcrackhome Mar 20 '13 at 2:06
  • $\begingroup$ @Sebastian: You’re welcome. $\endgroup$ – Brian M. Scott Mar 20 '13 at 2:14
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If you take a $u$ that was $dv$ before you will get back where you started. The idea is to exploit the fact that taking two derivatives of $\cos$ get you back something that looks like the original but has a minus sign. If you do $$\int u dv=uv-\int v du$$ then integrate by parts again you get $$\int u dv=uv-\int v du=uv-\left(uv-\int udv\right)$$ which is not very interesting.

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