How Can Reduction Formulas Simplify the Integral of \( x^m(1-x)^n \)?

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SUMMARY

The integral \(\int^{1}_{0}x^{m}(1-x)^{n} dx\) can be simplified to \(\frac{m!n!}{(m+n+1)!}\) for all integers \(m, n \geq 0\). The discussion highlights the use of reduction formulas, specifically the reduction formula \(I_{m, n} : \frac{m-1}{m+n}\) for \(m \geq 2\) and \(\frac{n-1}{m+n}\) for \(n \geq 2\). However, integration by parts is recommended as a more straightforward approach to solving this integral, as substitution methods may complicate the process unnecessarily.

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Students and educators in mathematics, particularly those focusing on calculus, as well as anyone interested in simplifying complex integrals using reduction formulas and integration techniques.

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1. The problem

Show that [tex]\int^{1}_{0}x^{m}(1-x)^{n} dx = \frac{m!n!}{(m+n+1)!}[/tex] for all integers [tex]m, n \geq 0[/tex]

The question is under "Reduction" topic, so I assume we solve this via reduction.

2. My attempt

My attempt is as follows:

Let [tex]x = cos^{2}x[/tex]

Then we get [tex]\frac{1}{2}\int cos^{2m-1}sin^{2n-1}dx[/tex]

From here I use the reduction formula: [tex]I_{m, n} : \frac{m-1}{m+n} : m \geq 2[/tex] or [tex]\frac{n-1}{m+n} : n \geq 2[/tex]

It seems like I am on the right track, but it's not working out properly. Am I missing something?
 
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don't really know how to help but...

wow. that looks like a monster problem! what hideousness of an equation that is!
 
Solve using integration by parts. You should get a redundant term or otherwise be able to simplify the problem in a few steps. The substitution method looks unnecessarily complicated here, and that too is based on integration by parts...so you'd be better off trying this problem from the basics up.
 

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