- #1
stripes
- 266
- 0
Homework Statement
should be my last question for at least the next few days...here goes...
Prove that, for even powers of sine,
[tex]\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}[/tex]
Homework Equations
[tex]
uv - \int v du = \int u dvdx
[/tex]
The Attempt at a Solution
let [itex]u = sin^{2n-1}x[/itex] and [itex]dv = sin x dx[/itex]
so [itex]du = (2n-1)(sin^{2n-2}x)(cos x)dx[/itex] and [itex]v = -cos x[/itex]
and we get:
[tex]\int sin^{2n}x dx = (-cos x)(sin^{2n-1}x) + (2n-1)\int (cos^{2}x)(sin^{2n-1}x)[/tex]
and i used integration by parts again
let [itex]u = sin^{2n-2}x[/itex] and [itex]dv = cos^{2}x dx[/itex]
so [itex]du = (2n-2)(sin^{2n-3}x)(cos x)dx[/itex] and [itex]v = \frac{1}{2}((sin x)(cos x) + x)[/itex]
then we get:
[tex]\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[/tex]
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: