Reduction Formula

**Simplicity** · April 24th 2008, 01:14 PM

Please help with this question. Thanks in advance.

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

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

**Mathstud28** · April 24th 2008, 01:19 PM

Originally Posted by Air

Please help with this question. Thanks in advance.

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

**Simplicity** · April 24th 2008, 01:23 PM

Originally Posted by Mathstud28

Use parts twice

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

**Mathstud28** · April 24th 2008, 01:25 PM

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

**Jhevon** · April 24th 2008, 01:27 PM

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)

**Danshader** · April 24th 2008, 01:33 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).

**Simplicity** · April 26th 2008, 09:06 AM

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.

**Moo** · April 26th 2008, 09:19 AM

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

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

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

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

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

**Simplicity** · April 26th 2008, 09:33 AM

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

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

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

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

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.

**Moo** · April 26th 2008, 09:35 AM

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

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