# Math Help - 'Reduction Formula' problem

1. ## 'Reduction Formula' problem

I'm just not sure how to go about solving this:

In= ∫(x^n).(1-x)^(3/2)dx
The integral has limits from 1 -> 0

And I need to show :

In = (2n/2n+5) I(n-1)

Thanks in advance for any help

2. ## Re: 'Reduction Formula' problem

Originally Posted by Mrhappysmile
I'm just not sure how to go about solving this:

In= ∫(x^n).(1-x)^(3/2)dx
The integral has limits from 1 -> 0

And I need to show :

In = (2n/2n+5) I(n-1)
Integrate by parts, using the fact that $\int(1-x)^{3/2} = -\tfrac25(1-x)^{5/2}$. Then

\begin{aligned}I_n &= \Bigl[-\tfrac25(1-x)^{5/2}x^n\Bigr]_0^1 + \int_0^1nx^{n-1}\tfrac25(1-x)^{5/2}dx \\ &= \tfrac{2n}5\int_0^1 x^{n-1}(1-x)(1-x)^{3/2}dx = \tfrac{2n}5I_{n-1} - \tfrac{2n}5I_n. \end{aligned}

3. ## Re: 'Reduction Formula' problem

Originally Posted by Opalg
Integrate by parts, using the fact that $\int(1-x)^{3/2} = -\tfrac25(1-x)^{5/2}$. Then

\begin{aligned}I_n &= \Bigl[-\tfrac25(1-x)^{5/2}x^n\Bigr]_0^1 + \int_0^1nx^{n-1}\tfrac25(1-x)^{5/2}dx \\ &= \tfrac{2n}5\int_0^1 x^{n-1}(1-x)(1-x)^{3/2}dx = \tfrac{2n}5I_{n-1} - \tfrac{2n}5I_n. \end{aligned}
Of course! I'd done exactly what you'd done but I didn't seen the simple bit i.e. dividing through by $5+2n$ to give the answer.

Thanks