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How to integrate 0 to 1 (1x^2)^n 
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#1
Nov1612, 10:42 AM

P: 2

1. The problem statement, all variables and given/known data
The reduction formula is: [tex] \int (1x^2)^n dx = (1x^2)^n x + 2n \int x^2(1x^2)^{n1} dx [/tex] and the question is: use this formula above how many times is necessary to prove: [tex] \int^{1}_{0} (1x^2)^n dx = 2n \frac{2(n1)}{3} \frac{2(n2)}{5} ... \frac{4}{2n3} \frac{2}{(2n1)(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 (1x^2)^n dx = (1x^2)^n x + 2n \int x^2(1x^2)^{n1} dx [/tex] Integrating by parts: [tex] =x(1x^2)^n+2n \int x^2(1x^2)^{n1}dx [/tex] let [tex] x^2 = (1x^2)+1 [/tex] [tex]=x(1x^2)^n2n \int(1x^2)^n dx + 2n \int (1x^2)^{n1}dx [/tex] but I still don't know how to get there with this formula. Any guidance would be appreciated. 


#2
Nov1612, 11:12 AM

P: 206

use your last equality and induction.see attached.



#3
Nov1612, 11:20 AM

P: 3,096

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
Nov1612, 02:49 PM

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How to integrate 0 to 1 (1x^2)^n
[itex]\displaystyle \int (1x^2)^n dx=x(1x^2)^n2n \int(1x^2)^n dx + 2n \int (1x^2)^{n1}dx\ .[/itex] That has [itex]\displaystyle \ \int (1x^2)^n dx\ \ [/itex] on both sides of the equation. Solve for [itex]\displaystyle \ \int (1x^2)^n dx\ .[/itex] 


#5
Nov1912, 03:55 PM

P: 2

How did you get this formula? 


#6
Nov1912, 04:30 PM

P: 206

substitute the limits in your last equality.



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