Prove Reduction Formula: I_n = 3n/(3n+1)I_{n-1}

[subzero0137]

If $I_{n}=\int_0^1 (1-x^{3})^{n} dx$, use integration by parts to prove the reduction formula $I_{n}=\frac{3n}{3n+1}I_{n-1}$My attempt: let $u=(1-x^{3})^{n}$, and $dv=dx$. Then $I_{n}=[(1-x^{3})^{n}x]_0^1 - \int_0^1 -3x^{2}n(1-x^{3})^{n-1}x dx = 3n \int_0^1 x^{2}(1-x^{3})^{n-1} dx$. But I don't know where to go from here. Any help would be appreciated.

 

[Dick]

If $I_{n}=\int_0^1 (1-x^{3})^{n} dx$, use integration by parts to prove the reduction formula $I_{n}=\frac{3n}{3n+1}I_{n-1}$


My attempt: let $u=(1-x^{3})^{n}$, and $dv=dx$. Then $I_{n}=[(1-x^{3})^{n}x]_0^1 - \int_0^1 -3x^{2}n(1-x^{3})^{n-1}x dx = 3n \int_0^1 x^{2}(1-x^{3})^{n-1} dx$. But I don't know where to go from here. Any help would be appreciated.

You have an x^3 in your last integral. Not an x^2. Here's a hint. Try writing x^3=(1-x^3)-1.

 

[subzero0137]

You have an x^3 in your last integral. Not an x^2. Here's a hint. Try writing x^3=(1-x^3)-1.

After writing x^3=(1-x^3)-1, should I integrate by parts again?

 

[Dick]

After writing x^3=(1-x^3)-1, should I integrate by parts again?

I think you can think of something cleverer than that. Split it into two integrals and take a close look at them.

 

[Dick]

After writing x^3=(1-x^3)-1, should I integrate by parts again?

Ooops. I've got a typo. That should obviously be x^3=(x^3-1)+1. Sorry!

 

[subzero0137]

Ooops. I've got a typo. That should obviously be x^3=(x^3-1)+1. Sorry!

I've got it now. Thanks :D