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


by mecattronics
Tags: 1x2n, integrate
mecattronics
mecattronics is offline
#1
Nov16-12, 10:42 AM
P: 2
1. The problem statement, all variables and given/known data

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

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

[tex] \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)}[/tex]
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:

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

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

Any guidance would be appreciated.
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hedipaldi
hedipaldi is offline
#2
Nov16-12, 11:12 AM
P: 206
use your last equality and induction.see attached.
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002.jpg  
jedishrfu
jedishrfu is offline
#3
Nov16-12, 11:20 AM
P: 2,493
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.

SammyS
SammyS is offline
#4
Nov16-12, 02:49 PM
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How to integrate 0 to 1 (1-x^2)^n


Quote Quote by mecattronics View Post
1. The problem statement, all variables and given/known data

The reduction formula is:[tex] \int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx [/tex]and the question is:
use this formula above how many times is necessary to prove:[tex] \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)}[/tex]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:[tex] \int (1-x^2)^n dx = (1-x^2)^n x + 2n \int x^2(1-x^2)^{n-1} dx [/tex]Integrating by parts:
[tex] =x(1-x^2)^n+2n \int x^2(1-x^2)^{n-1}dx [/tex]
let [tex] x^2 = -(1-x^2)+1 [/tex]
[tex]=x(1-x^2)^n-2n \int(1-x^2)^n dx + 2n \int (1-x^2)^{n-1}dx [/tex]
but I still don't know how to get there with this formula.

Any guidance would be appreciated.
Your last line says that

[itex]\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\ .[/itex]

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

Solve for [itex]\displaystyle \ \int (1-x^2)^n dx\ .[/itex]
mecattronics
mecattronics is offline
#5
Nov19-12, 03:55 PM
P: 2
Quote Quote by hedipaldi View Post
use your last equality and induction.see attached.
Hi hedipaldi, thanks for your help!

How did you get this formula?
hedipaldi
hedipaldi is offline
#6
Nov19-12, 04:30 PM
P: 206
substitute the limits in your last equality.


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