1. ## Reduction Formulae

I need some help on this reduction formulae question.

The question is:
Given that $\displaystyle \ I_n = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx \$
Show that
$\displaystyle \ I_{2n} = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }} \$

I know that the reduction formulae is
$\displaystyle \ I_n = \frac{{n - 1}}{n}I_{n - 2} \$
But I don't know how manipulate it into the above form. Could someone help please?

2. Originally Posted by free_to_fly
I need some help on this reduction formulae question.

The question is:
Given that $\displaystyle \ I_n = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx \$
Show that
$\displaystyle \ I_{2n} = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }} \$

I know that the reduction formulae is
$\displaystyle \ I_n = \frac{{n - 1}}{n}I_{n - 2} \$
But I don't know how manipulate it into the above form. Could someone help please?
$\displaystyle I_0 = \int_0^{\pi /2} sin^0(x)dx = \int_0^{\pi /2} dx = [x]_0^{\pi /2} = \pi /2$

$\displaystyle I_1 = \int_0^{\pi /2} sin^1(x)dx = \int_0^{\pi /2} sin(x) dx = [-cos(x)]_0^{\pi /2} = 1$

-----
n=1
$\displaystyle I_2 = \frac 12 * \frac{\pi}2$

-----
n=2
$\displaystyle I_4 = \frac 34 * \frac 12 * \frac{\pi}2$

-----
n=3
$\displaystyle I_6 = \frac 56 *\frac 34 * \frac 12 * \frac{\pi}2$

-----
n=4
$\displaystyle I_8 = \frac 78 * \frac 56 *\frac 34 * \frac 12 * \frac{\pi}2$

By now you should be noticing the pattern.

Lets look at the numerator:
the leading term is always 2n-1 and it skips every other number. so we can write it like this:

$\displaystyle 7*5*3*1*\pi$

$\displaystyle = \frac{8*7*6*5*4*2*1}{8*6*4*2}\pi$

$\displaystyle =\frac{8!}{4(2)*3(2)*2(2)*1(2)}\pi$

$\displaystyle = \frac{8!}{(2*2*2*2)(4*3*2*1)}\pi$

$\displaystyle = \frac{8!}{2^4*4!}\pi$

$\displaystyle = \frac{(2*4)!}{2^4*4!}\pi$

And of course you notice that this n=4 and that this really means the numerator is

$\displaystyle = \frac{(2*n)!}{2^n*n!}\pi$

-----

And now we look at the denominator
$\displaystyle 8*6*4*2*2$ And we notice that the leading term is always 2n. So we evaluate it a bit.

$\displaystyle 8*6*4*2*2$

$\displaystyle =4(2)*3(2)*2(2)*2(1)*2$

$\displaystyle =(2*2*2*2)(4*3*2*1)*2$

$\displaystyle =2^4*4!*2$

$\displaystyle =(2*2^4)*4!$

$\displaystyle =2^{4+1}*4!$

And of course you notice that n=4 and that this holds for any iteration. So the denominator becomes

$\displaystyle =2^{n+1}*n!$

-----

Then you consolidate the function for the numerator and denominator
$\displaystyle \frac{\frac{(2*n)!}{2^n*n!}\pi}{2^{n+1}*n!}$

$\displaystyle =\frac{(2*n)!}{2^n*n!}\pi * \frac 1{2^{n+1}*n!}$

$\displaystyle =\frac{(2n)!\pi}{2^n*n!*2^{n+1}*n!}$

$\displaystyle =\frac{(2n)!\pi}{(2^n*2^{n+1})(n!*n!)}$

$\displaystyle =\frac{(2n)!\pi}{2^{n+1+n}(n!)^2}$

$\displaystyle =\frac{(2n)!\pi}{2^{2n+1}(n!)^2}$

3. Originally Posted by free_to_fly
I need some help on this reduction formulae question.

The question is:
Given that $\displaystyle \ I_n = \int_0^{\frac{\pi }{2}} {\sin ^n } xdx \$
Show that
$\displaystyle \ I_{2n} = \frac{{(2n)!\pi }}{{2^{2n + 1} (n!)^2 }} \$

I know that the reduction formulae is
$\displaystyle \ I_n = \frac{{n - 1}}{n}I_{n - 2} \$
But I don't know how manipulate it into the above form. Could someone help please?