How to integrate 0 to 1 (1-x^2)^n

1. Nov 16, 2012

mecattronics

1. The problem statement, all variables and given/known data

The reduction formula is:
$$\int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx$$

and the question is:
use this formula above how many times is necessary to prove:

$$\int^{1}_{0} (1-x^2)^n dx = 2n \frac{2(n-1)}{3} \frac{2(n-2)}{5} ... \frac{4}{2n-3} \frac{2}{(2n-1)(2n+1)}$$
but I don't know how to get there.

2. Relevant equations
-

3. The attempt at a solution
I tried to modify the reduction formula leaving it more recursive:

$$\int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx$$
Integrating by parts:
$$=x(1-x^2)^n+2n \int x^2(1-x^2)^{n-1}dx$$
let $$x^2 = -(1-x^2)+1$$
$$=x(1-x^2)^n-2n \int(1-x^2)^n dx + 2n \int (1-x^2)^{n-1}dx$$

but I still don't know how to get there with this formula.

Any guidance would be appreciated.

2. Nov 16, 2012

hedipaldi

use your last equality and induction.see attached.

Attached Files:

• 002.jpg
File size:
11.3 KB
Views:
231
3. Nov 16, 2012

Staff: Mentor

Since you have a definite integral perhaps you could tryparametric differentiation to get the answer.

The only reference I have for this is Prof Nearings Mathematical Methods pdf at:

http://www.physics.miami.edu/~nearing/mathmethods/

section 1.2 pg 4.

4. Nov 16, 2012

SammyS

Staff Emeritus

$\displaystyle \int (1-x^2)^n dx=x(1-x^2)^n-2n \int(1-x^2)^n dx + 2n \int (1-x^2)^{n-1}dx\ .$

That has $\displaystyle \ \int (1-x^2)^n dx\ \$ on both sides of the equation.

Solve for $\displaystyle \ \int (1-x^2)^n dx\ .$

5. Nov 19, 2012

mecattronics

Hi hedipaldi, thanks for your help!

How did you get this formula?

6. Nov 19, 2012

hedipaldi

substitute the limits in your last equality.