Integration reduction formula

[Petrushka]

I'm trying to find an integration reduction formula for the following equation:


{{I}_n}=\int _{0}^{2}{{\big(4-{x^2}\big)}^n}\delta x


Any indication on how to begin would be much appreciated as I've tried many different approaches but all have ended in failure.

Thanks

[Tide]

I'd try to repeatedly integrate by parts or possibly use the binomial expansion.

[Zurtex]

I think I have a solution just give me 5 mins to see if it works.

[Zurtex]

O.K it's been quite a few months since I've done this, so I can't remember if this reduction formulae is fairly simple.

If you use the substitution:

x = 2 \sin u

It becomes:

\frac{1}{2}4^n\int_0^{\frac{\pi}{2}} \left( \cos^{2n-1} u \right) du

I'm sure that can be done with a few trig identities and standard results but it's too late for me to think about it sorry.

[Petrushka]

Thanks for the swift responses guys I'll have a go at that tomorrow.

[Zurtex]

I'm not sure about this but referring back to my previous post could you just let m = 2n - 1 for n > 0 and then that's a fairly standard reduction formulae. I've never done something like that for a reduction formulae but I don't see why it can't be done.

[Petrushka]

You may well be able to do that, Zurtex, and I also don't see why it wouldn't work, the only trouble is that it wouldn't prove the relation I was asked to prove.

I managed to solve it (with help from maths teacher) using a very clever trick indeed. The solution is as follows if anyone is interested:


{I }_n}\multsp =\int _{0}^{2}{{\big(4-{x^2}\big)}^n}\delta x \\\noalign\vspace{1.08333ex}} \\= {{{{\big[x{{\big(4-{x^2}\big)}^n}\big]}_0}}^2}+2n\int _{0}^{2}{x^2}{{\big(4-{x^2}\big)}^{n-1}}\delta
x \\\noalign{\vspace{1.08333ex}} \\ \multsp \multsp \multsp \multsp \multsp \multsp =\multsp 2n\int _{0}^{2}\big(4-\big[4-{x^2}\big]\big){{\big(4-{x^2}\big)}^{n-1}}\delta x



\noalign{\vspace{1.08333ex}} \\ {{I }_n}\multsp \multsp =\multsp 8n\int _{0}^{2}{{\big(4-{x^2}\big)}^{n-1}}\delta x-2n\multsp {{I }_n} \\ \noalign{\vspace{0.833333ex}}



{{I }_n}\multsp \multsp =\multsp 8n\multsp {{I }_{n-1}}-2n\multsp {{I }_n}



\noalign{\vspace{0.916667ex}} \\
{{I }_n}\multsp \multsp =\multsp \frac{8n}{2n+1}{{I }_{n-1}


The trick, which I wouldn't have thought of for a very long time, was to write the x^2 term as (4-[4-x^2])

[Zurtex]

I thought I'd seen that before, that's really silly of me not to spot. Well done for working it out.

[Nebula]

What happens in the very first step of the solution?