# Math Help - Reduction Formula

1. ## Reduction Formula

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

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

2. 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

3. Originally Posted by Mathstud28
Use parts twice
Which should I take as $u$ and $\frac{\mathrm{d}v}{\mathrm{d}x}$ to be?

4. 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

5. 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)

6. 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).

7. 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.

8. 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 ?

9. 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.

10. 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