Reduction Formula

**bobak** · March 22nd 2008, 04:03 PM

Given that $I_{n} = \int_{0}^{1}{x^n(1-x)^{\frac{1}{2}}dx}$ where $n \geq 0$ , prove that for $n \geq 1$

$I_{n} = \frac{2n}{2n+3} I_{n-1}$

The first thing that comes to mind is $x = \sin^2u$ so $dx = 2 \sin u \cos u du$

giving $I_{n} = \int_{0}^{\frac{\pi}{2}}{sin^{2n} \cos u \ 2 \sin u \cos u \ du }$

$\Rightarrow 2 \int_{0}^{\frac{\pi}{2}}{sin^{2n+1} \cos^2 u \ du }$

my first thought is to use integration by part, with $du = sin^{2n+1} \cos u$ and $v = \cos u$

this doesn't get me very far sadly....

**TwistedOne151** · March 22nd 2008, 05:00 PM

Just do integration by parts without the substitution:
$u=x^n$ , $dv=\sqrt{1-x}dx$
Thus $du=nx^{n-1}dx$ , $v=-\frac{2}{3}(1-x)^{\tfrac{3}{2}}$
and:
$I_{n}=\int_{0}^{1}x^n\sqrt{1-x}\,dx$
$I_{n}=-\frac{2}{3}x^n(1-x)^{\tfrac{3}{2}}|_{x=0}^1-\int_{0}^{1}(nx^{n-1})\cdot\left(-\frac{2}{3}(1-x)^{\tfrac{3}{2}}\right)\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)^{\tfrac{3}{2}}\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}x^{n-1}(1-x)\sqrt{1-x}\,dx$
$I_{n}=\frac{2n}{3}\int_{0}^{1}(x^{n-1}-x^n)\sqrt{1-x}\,dx$
$I_{n}=\frac{2n}{3}\left(\int_{0}^{1}x^{n-1}\sqrt{1-x}\,dx-\int_{0}^{1}x^n\sqrt{1-x}\,dx\right)$
$I_{n}=\frac{2n}{3}(I_{n-1}-I_{n})$
and algebra can do the rest.

--Kevin C.

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