HELP!!!! I need to prove this!!!!!!!!!!!!!!!!!!?

Question

How do you prove this:

(-1)^n ∫ [-1, 1] (x² - 1)^n dx = [(2^(2n + 1))(n!)²]/(2n + 1)!

Please help me. I've tried doing it by induction but I have gotten anywhere.

Thanks!

Answer 1

Take a breath and calm down!!!! You'll have it proven!!!!

(-1)^n ∫ [-1, 1] (x² - 1)^n dx = ∫ [-1, 1] (1 - x²)^n dx =

/even function/

= 2 * ∫ [0, 1] (1 - x²)^n dx =

/now take x = sin t, dx = cos t dt, x=0 → t=0, x=1 → t=π/2/

= 2 * ∫ [0, π/2] (cos t)^(2n+1) dt

Let I_{2n+1} = ∫ [0, π/2] (cos t)^(2n+1) dt, integrate by parts now to derive a recurrence relationship between I_{2n+1} and I_{2n-1}, allowing to bring the things down to

I_{1} = ∫ [0, π/2] cos t dt = 1

I_{2n+1} = ∫ [0, π/2] (cos t)^(2n) d(sin t) =

= [(cos t)^(2n)*sin t][0, π/2] - ∫ [0, π/2] (sin t) d(cos t)^(2n) =

= 2n * I_{2n-1} - (2n-1) * I_{2n+1},

so the recurrence relationship is

I_{2n+1} = (2n/(2n+1)) * I_{2n-1}

Using it necessary number of times:

I_{2n+1) = 2n*(2n-2)* . . *4*2 / (2n+1)(2n-1)* . . *3*1 = (2n)!!/(2n+1)!!,

hence the answer is

2 * (2n)!!/(2n+1)!! = 2 * ((2n)!!)² / (2n+1)! =

= 2^(2n+1) * (n!)² / (2n+1)! as required.

Answer 2

OK, here goes:

First, let's write

g(n) =  [(2^(2n + 1))(n!)²]/(2n + 1)!

Also note that g(n) = 4n² / (2n(2n+1)) * g(n-1),

which simplifies to

g(n) = 2n / (2n+1) * g(n-1).

So once you verify it for n=1, to complete the induction, you need to show the above relation.  So we integrate by parts, letting

u = (1-x²)^n, and

v = x.

We get

∫ (1-x²)^n dx = x(1-x²)^n [-1 to 1] + 2n∫ x²(1-x²)^(n-1) dx.

This implies

∫ (1-x²)^(n-1) dx = (2n+1) ∫ x²(1-x²)^(n-1) dx.

Using induction, we get

∫ x²(1-x²)^(n-1) dx = g(n-1) / (2n+1).

So

∫ (1-x²)^n dx = ∫ (1-x²)^(n-1) dx - ∫ x²(1-x²)^(n-1) dx = g(n-1) - g(n-1) / (2n+1) = g(n-1) * 2n / (2n+1).

Steve

Answer 3

you need to prove this with mathematical induction. you probably just made a mistake. try it again.

Answer 4

wow ;(

im lost

Answer 5

sorry its lost me