# Proof of reduction formula for a definite integral of an even powered sine function

1. Jun 3, 2010

### stripes

1. The problem statement, all variables and given/known data
should be my last question for at least the next few days...here goes...

Prove that, for even powers of sine,

$$\int^{\frac{\pi}{2}}_{0}sin^{2n}x dx = \frac{2\cdot4\cdot6\cdot...\cdot(2n - 1)}{2\cdot4\cdot6\cdot...\cdot2n}\cdot\frac{\pi}{2}$$

2. Relevant equations

$$uv - \int v du = \int u dvdx$$

3. The attempt at a solution

let $u = sin^{2n-1}x$ and $dv = sin x dx$
so $du = (2n-1)(sin^{2n-2}x)(cos x)dx$ and $v = -cos x$

and we get:

$$\int sin^{2n}x dx = (-cos x)(sin^{2n-1}x) + (2n-1)\int (cos^{2}x)(sin^{2n-1}x)$$

and i used integration by parts again

let $u = sin^{2n-2}x$ and $dv = cos^{2}x dx$
so $du = (2n-2)(sin^{2n-3}x)(cos x)dx$ and $v = \frac{1}{2}((sin x)(cos x) + x)$

then we get:

$$\int sin^{2n}x dx = (-cos x)(sin^{2n-1}x) + \frac{2n-1}{2}(sin^{2n-2}(sinx cosx + x)) - (2n-2)\int ((sinx cosx + x)(sin^{2n-3}x)(cos x))dx$$

now i've realized i'm just pointlessly integrating by parts over and over...it's just getting harder and harder (and more difficult to put here on PF!)

If someone could guide me in the right direction for proving this formula, I would appreciate it. Thank you so much in advance!

Last edited: Jun 3, 2010
2. Jun 3, 2010

### System

Re: proof of reduction formula for a definite integral of an even powered sine functi

Did you forget that you have a definite integral?

3. Jun 3, 2010

### HallsofIvy

Re: proof of reduction formula for a definite integral of an even powered sine functi

In addition to remembering that this is a definite integral, so that you don't have that long sum, I would try proof by induction so that I only have to do the integration by parts once.

4. Jun 3, 2010

### stripes

Re: proof of reduction formula for a definite integral of an even powered sine functi

i knew all along that we have a definite integral, but we still need to get rid of that integral sign on the end of the whole thing, there'll always be that last integral...

our course hasn't taught us enough to integrate that...