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

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Homework Statement



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.

Homework Equations


-

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.
 
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use your last equality and induction.see attached.
 

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mecattronics said:

Homework Statement



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

Homework Equations



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} dxIntegrating 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.
Your last line says that

\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\ .
 
hedipaldi said:
use your last equality and induction.see attached.

Hi hedipaldi, thanks for your help!

How did you get this formula?
 
substitute the limits in your last equality.
 

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