# Reduction Formula

• Apr 24th 2008, 02:14 PM
Simplicity
Reduction Formula

Q:
$I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

Prove that $I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2$.
• Apr 24th 2008, 02:19 PM
Mathstud28
Quote:

Originally Posted by Air

Q:
$I_n = \displaystyle\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

Prove that $I_n = \frac{\pi ^n}{2^n} - n (n - 1) I_{n-2}, \ n \geq 2$.

Use parts twice
• Apr 24th 2008, 02:23 PM
Simplicity
Quote:

Originally Posted by Mathstud28
Use parts twice

Which should I take as $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to be?
• Apr 24th 2008, 02:25 PM
Mathstud28
Quote:

Originally Posted by Air
Which should I take as $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to be?

I would say cos(x) for u since $\frac{D[\cos(x)]^2}{dy^2}=-\cos(x)$ which seems to be a part of the second term in the proof
• Apr 24th 2008, 02:27 PM
Jhevon
Quote:

Originally Posted by Air
Which should I take as $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to be?

u = x^n and du/dx = cos(x) seems to be the right choice. (the n(n - 1) part of the formula hints at this)
• Apr 24th 2008, 02:33 PM
generally i will go for u = x^n and dv/dt = cos x. My reasoning is that if x^n is taken as dv/dt then the power of x will increase which brings you nowhere closer in solving the integral. However if you set u = x^n for each part the power of x decreases till x^0 (this gives an integration of a trigonometry which is simple to solve).
• Apr 26th 2008, 10:06 AM
Simplicity
Thanks all.

I've done it and obtained the correct answer however my method was pretty long and slightly doubtful. Could someone also do this question so I can see if they do this 'Reduction Formula' method. I will be grateful if someone could help. Thanks in advance.
• Apr 26th 2008, 10:19 AM
Moo
Hello,

I don't think you could avoid it...

$I_n =\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

I'll name $\frac{\pi}{2}$ y because it's really awful to write :D

$=[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx$

$=y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}$

$=\frac{\pi^n}{2^n}+n(n-1)I_{n-2}$

Is it your reduction formula or are you looking for the induction thing ?
• Apr 26th 2008, 10:33 AM
Simplicity
Quote:

Originally Posted by Moo
Hello,

I don't think you could avoid it...

$I_n =\int^\frac{\pi}{2}_0 x^n \cos x \, \mathrm{d}x$

I'll name $\frac{\pi}{2}$ y because it's really awful to write :D

$=[-x^n \sin x]_0^y + n \int_0^y x^{n-1} \sin x dx$

$=y^n+n \underbrace{\left( [x^{n-1} \cos x]_0^y \right)}_{=0} + n(n-1) \underbrace{\int_0^y x^{n-2} \cos x dx}_{I_{n-2}}$

$=\frac{\pi^n}{2^n}+n(n-1)I_{n-2}$

Is it your reduction formula or are you looking for the induction thing ?

No, it's the method you've done. I tend to write each stage so the method appears long but how you have done it has helped clarify the method. Thanks. (Smile)
• Apr 26th 2008, 10:35 AM
Moo
I know I've missed some steps, but it should add one or two more lines for each line..
I hope this will really help (Bow)