1. ## 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.

2. 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

3. 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$$\displaystyle - (n-1)\int sin^nxdx \displaystyle I_n = -sin^{n-1}xcosx + (n-1)I_{n-2} - (n-1)I_n \displaystyle nI_n = -sin^{n-1}xcosx + (n-1)I_{n-2} \displaystyle I_n = \frac{-sin^{n-1}xcosx + (n-1)I_{n-2}}{n} See if you can do the rest 4. Originally Posted by Random333 Ok, I think I've worked out your explanation. \displaystyle \int^{\frac {\pi}{2}}_{\frac{\pi}{4}} \sin^{n}x dx Integral in terms of x using n is \displaystyle [\frac {-\cos^{n+1}x}{n+1}]^{\frac {\pi}{2}}_{\frac{\pi}{4}} And on you go, hopefully....if I can work LaTeX This I'm afraid is not correct 5. Hello, sterps! This is a Reduction Formula derivation. It's quite long and elaborate . . . better sit down. Use integration by parts to express: \displaystyle I(n) = \int \sin^n\!x\\, dx in terms of \displaystyle I(n - 2). Given: .\displaystyle I(n)\;=\;\int\sin^n\!x\,dx Let: \displaystyle u = \sin^{n-1}\!x\qquad\qquad\qquad\qquad dv = \cos x\,dx Then: \displaystyle du = (n-1)\sin^{n-2}\!x\cos x\,dx\qquad v = -\sin x We have: .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x - \int[-\cos x][(n-1)\sin^{n-1}\!x\cdot\cos x\,dx . . . . . . . .\displaystyle I(n)\;= \;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x\cos^2\!x\,dx . . . . . . . .\displaystyle I(n) \;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\sin^{n-2}\!x(1 - \sin^2\!x)\,dx . . . . . . . .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\!\int\left[\sin^{n-2}\!x - \sin^n\!x\right]\,dx . . . . . . . .\displaystyle I(n)\;=\;-\sin^{n-1}\!x\cos x + (n-1)\int\sin^{n-2}\!x\,dx -$$\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 . . .

6. thanks for the help guys.

ive never heard of Reduction Formula derivation.

7. Well it is the name of what you are trying to do.

Kind of turning an integration problem into an iterative problem