Help to prove a reduction formula?

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Homework Statement



Let In = \int^{1}_{-1} (1-x^{2})^{n} dx.

Use integration by parts to show that In = (\frac{2n}{2n+1}) In-1 for n \geq1.

(The integral above is supposed to be between the limits -1 and 1... sorry I couldn't figure out how to make the limits appear properly.)

Homework Equations



Integration by parts.

The Attempt at a Solution



I have tried several different ways.

First of all I tried letting u=(1-x^{2})^{n} and dv/dx = 1. Hence, du/dx = (1-x^{2})^{n-1}(-2x) and v=x. Using this I got In = n \int^{1}_{-1} 2x2(1-x^{2})^{n-1} dx. This didn't seem to be very helpful.

I then tried writing In as \int^{1}_{-1} (1+x)^{n}(1-x)^{n} dx, and letting u be (1+x)^{n} and dv/dx be (1-x)^{n}. I thought this was a pretty clever idea, but it didn't give me what I wanted. I got In = \frac{n}{n+1} \int^{1}_{-1} (1-x2)n-1(1-x)2 dx, which can be written equivalently as \frac{n}{n+1} \int^{1}_{-1} (1-x)n+1(1+x)n-1 dx. I then tried doing another integration by parts, letting u=(1+x)n-1 and dv/dx=(1-x)n+1, and this gave me In = (\frac{n}{n+1})(\frac{n-1}{n+2}) \int^{1}_{-1} (1-x2)n-2(1-x)4 dx, or equivalently (\frac{n}{n+1})(\frac{n-1}{n+2}) \int^{1}_{-1} (1+x)n-2(1-x)n+2 dx , but I still didn't seem to be any closer.

The other thing I tried was writing In = \int^{1}_{-1} (\sqrt{1-x^2)2n dx, and letting u = (\sqrt{1-x^2})2n and dv/dx = 1. This gave me a slightly more interesting result of In = 2n \int^{1}_{-1} x2 (\sqrt{1-x^2})2n-2 dx , i.e. In = 2n \int^{1}_{-1} x2 (1-x^2})n-1 dx, but I wasn't sure what to do next.

Please help... what should I be doing?
 
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I think it would make a lot more sense to let u= 1-x2 and dv= (1- x2)n-1dx.
 
HallsofIvy said:
I think it would make a lot more sense to let u= 1-x2 and dv= (1- x2)n-1dx.


But then I would have to integrate (1- x2)n-1dx, and I don't know how to do that when n is unknown. Am I missing something? I tried to integrate it using MAPLE, and it gave me something useless to do with the hypergeometric function.
 
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