Recurrence question (1 Viewer)

[tarsus]

From Cambridge: Excercise 5.5, question 2:

In: ∫x(1-x^3)^n dx (limits: 1 and 0)

And show that: In = 3n/(3n+2) In-1


Here's a pic:

I did it by parts using u = (1-x^3)^n; u' = n(1-x^3)^(n-1) -(3x^2) dx; dv = x dx; v = (x^2)/2 and then I got screwed by the algebra.

The best I can do afterwards is break the -x^4 apart into x(-x^3) but don't know what to do afterwards...

 

[pLuvia]

int.x(1-x³)ⁿdx
u=(1-x³)ⁿ dv=dx
du=-3x²n(1-x³)^n-1 v=x

I_n=[1/2x²(1-x³)ⁿ]{0 to 1}-n/2 int.{0 to 1}x²(1-x³)^n-1dx
=3/2n int.{0 to 1}x⁴(1-x²)^n-1dx
=3/2n int.{0 to 1}x-x(1-x³)(1-x³)^n-1dx
=3/2n int.{0 to 1}x(1-x³)^n-1-x(1-x³)ⁿdx
=3/2n[I_n-1-I_n]
I_n=3/2nI_n-1-3/2nI_n
I_n=[3n/(3n+2)]I_n-1
QED