I know how to integrate, for example $x^2\sin x$ via integration by parts. But how would one approach $x^n\sin x$, where $n$ is an arbitrary natural number? Do I have have to use integration by parts $n$ times? How do I do this?
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1Do it by parts twice and you will then have a recurrence relation between $I_n$ and $I_{n-2}$, I bet. – Claude Leibovici Jun 28 '14 at 13:55
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And since you know the first two terms, yo are done. You can check since you know the result for $n=2$. – Claude Leibovici Jun 28 '14 at 14:05
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This answer might be useful: http://math.stackexchange.com/questions/70974/lesser-known-integration-tricks/71195#71195 – Hans Lundmark Jun 28 '14 at 14:17
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Integrating two times by parts one can find the recursive relation $$\begin{array}{rcl} I_{n}(t)&=& \int^t x^n \sin(x)dx \\ &=& -t^{n}\cos(t)+n\int^t x^{n-1} \cos(x)dx\\ &=& -t^n\cos(t)+nt^{n-1}\sin(t)-n(n-1)I_{n-2}(t)\end{array}$$
Surb
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I did the integration by parts twice and I have the recursive relation. Now what do I do with it? Am supposed to solve it for n? – Adam Jun 28 '14 at 14:16
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You know $I_1$ and $I_2$, so there is no problem, you just apply the formula recursively until one of these terms appears (depending on the parity of $n$). Here's a general closed form: http://www.wolframalpha.com/input/?i=integrate+x%5Ensin%28x%29 – Surb Jun 28 '14 at 14:32