Prove Reduction Formula for Integral: x(1-x^3)^n dx

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Homework Statement



In= (the integral) x(1-x^3)^ndx

Prove that (3n +2)In = 3nIn-1 + x^2(1 - x)^n

Hence find In in terms of n

The Attempt at a Solution



I tried integration by parts (by letting u be (1-x^3)^n and got stuck after this:
In = 1/2x^2(1-x^3)^n + (the integral)(n(1-3x)^(n-1)(3x^2)(1/2x^2)dx)
= 1/2(x^2(1-x^3)^n +3n(the integral)(1-x^3)^(n-1)x^4dx

Would greatly appreciate any help and thanks in advance!
 
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I didn't find the result but I think you don't have the right thing to prove (the last x is an x^3).


(But I may be wrong).
 
Write I_n as integral of x*(1-x^3)^n=x*(1-x^3)*(1-x^3)^(n-1)=x(1-x^3)^(n-1)-x^4(1-x^3)^(n-1). You can express the pesky x^4(1-x^3)^(n-1) integral in terms of I_n and I_(n-1).
 
Ok, I get it now. Thanks(:
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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