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If I am asked to prove (that is exactly the phrasing, not 'establish') a reduction formula, e.g. for $$I_n=\int_0^1 x^n \cosh x~ \mathrm{d} x $$ that $$I_n=\sinh 1 - n\cosh 1 + n(n-1)I_{n-2}$$ where $n\geqslant0$, is it sufficient to just integrate by parts so that \begin{align*} I_n&=x^n \sinh x \bigg|_0^1 - n\int_0^1 x^{n-1}\sinh x~\mathrm{d}x\\&=\sinh1 - n\left[x^{n-1}\cosh x\bigg |_0^1 -(n-1)\int_0^1x^{n-2}\cosh x ~\mathrm{d}x \right]\\&=\sinh 1 -n\cosh1-n(n-1)I_{n-2}\end{align*} or do I have to prove it otherwise by induction?

Thank you.

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That's fine. I think you have to check $n=0,1$ technically. –  anon Dec 13 '12 at 2:23
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All done. Nothing to do. No induction needed or relevant. Proof complete for $n\ge 2$. –  André Nicolas Dec 13 '12 at 2:23
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Integrating by parts is enough, since you found the relation analytically. Induction should be used when you don't know or it's not possible to find exact analytical form but you have some hypothesis that you want to check. –  Kaster Dec 13 '12 at 2:25

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I would ask your instructor, but I would be content with just using IBP, thus not needing induction. Another thing - if it said to "verify", you could simply take the derivative of the expression they give for the reduction formula and you should get the original back since a function and its primitive are equal, up to a constant.

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