First Step in Finding an Integration Reduction Formula for (4-x^2)^n

[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?