# recursive integration formula

• Aug 30th 2006, 05:19 AM
sterps
need help with this integration by parts
I need help with this question

Use integration by parts to express:

I(n) = (int) sin^n(x) dx

in terms of I(n - 2). Hence show that:

(int) between pi/4 & pi/2 1/ sin^4(x) dx =4 / 3

help would be much appreciated

cheers.
• Aug 30th 2006, 06:03 AM
Random333
Ok, I think I've worked out your explanation.

$\displaystyle \int^{\frac {\pi}{2}}_{\frac{\pi}{4}} \sin^{n}x dx$

Integral in terms of x using n is

$\displaystyle [\frac {-\cos^{n+1}x}{n+1}]^{\frac {\pi}{2}}_{\frac{\pi}{4}}$

And on you go, hopefully....if I can work LaTeX
• Aug 30th 2006, 06:54 AM
Glaysher
Quote:

Originally Posted by sterps
I need help with this question

Use integration by parts to express:

I(n) = (int) sin^n(x) dx

in terms of I(n - 2). Hence show that:

(int) between pi/4 & pi/2 1/ sin^4(x) dx =4 / 3

help would be much appreciated

cheers.

$\displaystyle \int sin^n x dx = \int sin^{n-1}xsin x dx$

$\displaystyle u = sin^{n-1}x$ and $\displaystyle v' = sin x$

$\displaystyle u' = (n - 1)sin^{n-2}xcosx$ and $\displaystyle v = -cosx$

$\displaystyle Integral = -sin^{n-1}xcosx + (n-1)\int sin^{n-2}xcos^2xdx$

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

$\displaystyle Integral = -sin^{n-1}xcosx + (n-1)\int sin^{n-2}xdx$$\displaystyle - (n-1)\int sin^nxdx \displaystyle I_n = -sin^{n-1}xcosx + (n-1)I_{n-2} - (n-1)I_n \displaystyle nI_n = -sin^{n-1}xcosx + (n-1)I_{n-2} \displaystyle I_n = \frac{-sin^{n-1}xcosx + (n-1)I_{n-2}}{n} See if you can do the rest • Aug 30th 2006, 07:15 AM Glaysher Quote: Originally Posted by Random333 Ok, I think I've worked out your explanation. \displaystyle \int^{\frac {\pi}{2}}_{\frac{\pi}{4}} \sin^{n}x dx Integral in terms of x using n is \displaystyle [\frac {-\cos^{n+1}x}{n+1}]^{\frac {\pi}{2}}_{\frac{\pi}{4}} And on you go, hopefully....if I can work LaTeX This I'm afraid is not correct • Aug 30th 2006, 07:20 AM Soroban Hello, sterps! This is a Reduction Formula derivation. It's quite long and elaborate . . . better sit down. Quote: Use integration by parts to express: \displaystyle I(n) = \int \sin^n\!x\\, dx in terms of \displaystyle I(n - 2). Given: .\displaystyle I(n)\;=\;\int\sin^n\!x\,dx Let: \displaystyle u = \sin^{n-1}\!x\qquad\qquad\qquad\qquad dv = \cos x\,dx Then: \displaystyle du = (n-1)\sin^{n-2}\!x\cos x\,dx\qquad v = -\sin x We have: .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x - \int[-\cos x][(n-1)\sin^{n-1}\!x\cdot\cos x\,dx . . . . . . . .\displaystyle I(n)\;= \;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x\cos^2\!x\,dx . . . . . . . .\displaystyle I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x(1 - \sin^2\!x)\,dx . . . . . . . .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\left[\sin^{n-2}\!x - \sin^n\!x\right]\,dx . . . . . . . .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\int\sin^{n-2}\!x\,dx -$$\displaystyle (n-1)\!\underbrace{\int\sin^n\!x\,dx}$
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . This is $\displaystyle I(n)$

We have: .$\displaystyle I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x\,dx - (n-1)\!\cdot\!I(n)$

$\displaystyle I(n) + (n-1)\!\cdot\!I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\underbrace{\int\sin^{n-2}\!x\,dx}$

. . . . . . . . $\displaystyle n\!\cdot\!I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\cdot\!I(n-2)$

Therefore: .$\displaystyle \boxed{I(n) \;=\;-\frac{1}{n}\sin^{n-1}\!x\cos x + \frac{n-1}{n}\!\cdot\!I(n-2)}$

Whew! . . . I need a nap . . .

• Aug 30th 2006, 01:24 PM
sterps
thanks for the help guys.

ive never heard of Reduction Formula derivation. :confused:
• Aug 30th 2006, 11:47 PM
Glaysher
Well it is the name of what you are trying to do.

Kind of turning an integration problem into an iterative problem