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$j_n=\int_0^1 x^{2n}\sin(\pi x)dx$.

How do I show that $j_{n+1}= \frac{1}{\pi^2}\left[\pi-(2n+1)(2n+2)j_n\right]$?

I keep getting a recurring integration by parts and I can't simplify it.

Please show your working for me to see where I'm going wrong.

Many Thanks.

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1 Answer 1

Well, for the first partial integration we have $j_{n+1}=\int_{0}^1 x^{2n+2}\sin(\pi x)dx = [\frac{-1}{\pi}x^{2n+2}\cos(\pi x)]_0^1 + \frac{1}{\pi}(2n+2)\int_0^1 x^{2n-1}\cos(\pi x)dx$

The second one is similar, and if you choose your functions such that you differentiate $x^{2n+1}$, your desired expression should appear.

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