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I have seen cyclic integration by parts trick used for trignometric integrals. I got the impression that it is necessary for the trick to work that derivative of the function is cyclic. However it is not so, because the trick works for polynomials as well, for example integrate twice by parts $$\int_0^1x^2\times x^2 dx.$$ I would to know and understand more about the trick. For which pairs of functions the trick works and how many cycles it takes and can it be used for series also...

Bezdomnyi
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With $u=x^2,\ dv=x^2$ one has $du=2x,\ v=x^3/3$ so parts once will finish this. Specifically $uv-\int vdu=x^2\cdot x^3/3-\int (x^3/3)\cdot(2x)$ so that one gets a constant times $x^4$ in the remaining integral, and the integral has "cycled" back to itself in one use of parts. We arrive at $$\int x^4 = x^5/5-(2/3)\int x^4,$$ so that $(5/3)\int x^4=x^5/3$ and then $\int x^4=x^5/5.$

I think the "trick" of cycling works when the derivative of one factor times the integral of the other, or second derivative of one times second integral of second, etc, winds up only off by a constant from the original integral. [Hopefully the constant it is off by is not $1$ or else no result!]

coffeemath
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  • I calculated exactly like you did. – Bezdomnyi Mar 17 '15 at 08:56
  • @Bezdomnyi Well since it really only uses parts once, I thought I'd mention it since in your question you referred to using parts twice on this example. I think it's an intereating topic about cycling and parts. (upvoted the question) – coffeemath Mar 17 '15 at 09:02
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The infamous use of this trick is of course $$ I=\int \frac{dx}{x} = \int x \frac{dx}{x^2} = x\frac{-1}{x} + \int \frac{dx}{x} = 1+I, $$ which shows that you need to be careful with your integration constants.

Chappers
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