# recursive integration formula

• August 30th 2006, 06:19 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.
• August 30th 2006, 07:03 AM
Random333
Ok, I think I've worked out your explanation.

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

Integral in terms of x using n is

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

And on you go, hopefully....if I can work LaTeX
• August 30th 2006, 07:54 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.

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

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

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

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

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

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

$I_n = -sin^{n-1}xcosx + (n-1)I_{n-2} - (n-1)I_n$

$nI_n = -sin^{n-1}xcosx + (n-1)I_{n-2}$

$I_n = \frac{-sin^{n-1}xcosx + (n-1)I_{n-2}}{n}$

See if you can do the rest
• August 30th 2006, 08:15 AM
Glaysher
Quote:

Originally Posted by Random333
Ok, I think I've worked out your explanation.

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

Integral in terms of x using n is

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

And on you go, hopefully....if I can work LaTeX

This I'm afraid is not correct
• August 30th 2006, 08:20 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:

$I(n) = \int \sin^n\!x\\, dx$ in terms of $I(n - 2).$

Given: . $I(n)\;=\;\int\sin^n\!x\,dx$

Let: $u = \sin^{n-1}\!x\qquad\qquad\qquad\qquad dv = \cos x\,dx$

Then: $du = (n-1)\sin^{n-2}\!x\cos x\,dx\qquad v = -\sin x$

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

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

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

. . . . . . . . $I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\left[\sin^{n-2}\!x - \sin^n\!x\right]\,dx$

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

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

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

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

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

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

• August 30th 2006, 02:24 PM
sterps
thanks for the help guys.

ive never heard of Reduction Formula derivation. :confused:
• August 31st 2006, 12:47 AM
Glaysher
Well it is the name of what you are trying to do.

Kind of turning an integration problem into an iterative problem