# Thread: VERY HARD! Reduction Formula Question

1. ## VERY HARD! Reduction Formula Question

The question is as follows:

Given that $I_{n} = \int_0^\frac{\pi}{2} \sin^n (x) dx$ where $n \epsilon N$ (used capital N to show the sign for natural number or normal number, not too sure what it is)

a) show that $I_{2n} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2}$

b) show further that $I_{2n-1} I_{2n} = \frac{\pi}{4n}$

really need help with this one, its bugging me so much that i cant get it

2. Hello,
Originally Posted by Stylis10
The question is as follows:

Given that $I_{n} = \int_0^\frac{\pi}{2} \sin^n (x) dx$ where $n \epsilon N$ (used capital N to show the sign for natural number or normal number, not too sure what it is)

a) show that $I_{2n} = \frac{(2n)! \pi}{2^{2n+1} (n!)^2}$
So write $I_{2n}$ :

$I_{2n}=\int_0^{\pi/2} \sin^{2n} (x) ~dx$

But $\sin^{2n}(x)=\sin^{2n-2}(x) \sin^2(x)=\sin^{2n-2}(x)-\cos^2(x) \sin^{2n-2}(x)$

So $I_{2n}=I_{2n-2}-\int_0^{\pi/2} \cos^2(x) \sin^{2n-2}(x) ~dx \qquad {\color{red}\star}$

Now let's study $J=\int_0^{\pi/2} \cos^2(x) \sin^{2n-2}(x) ~dx$
Integrate by parts with $u=\cos(x) \Rightarrow du=-\sin(x)$ and $dv=\cos(x)\sin^{2n-2}(x) \Rightarrow v=\frac{1}{2n-1} \cdot \sin^{2n-1}(x)$

Hence we have :
$J=\left. \frac{1}{2n-1} \cos(x) \sin^{2n-1}(x) \right|_0^{\pi/2}+\frac{1}{2n-1} \int_0^{\pi/2} \sin^{2n}(x) ~dx$
$\boxed{J=\frac{1}{2n-1} \cdot I_{2n}}$

Now substitute in ${\color{red}\star}$ :

$I_{2n}=I_{2n-2}-\frac{1}{2n-1} \cdot I_{2n}$

Hence $I_{2n}=\frac{2n-1}{2n} \cdot I_{2(n-1)}$ (recursive formula)

Now it should be easy... And I voluntarily skipped steps so that you'll have to think a little

3. Thanxs Moo, but i got all of what you've done an upto that point, i however don't understand the factorails bit, its just eluding me

4. Originally Posted by Stylis10
Thanxs Moo, but i got all of what you've done an upto that point, i however don't understand the factorails bit, its just eluding me
You could have shown your work... I wouldn't have bothered explaining these things

So for the factorial thing, you can do it by induction.
If n=0, then $I_0=\pi/2$ but $\left. \frac{(2n)!\pi}{2^{2n+1}(n!)^2} \right|_{n=0}=\pi/2$. So it is ok.
Now assume $I_{2n}=\frac{(2n)!\pi}{2^{2n+1}(n!)^2}$

$I_{2(n+1)}=\frac{2(n+1)-1}{2(n+1)} \cdot I_{2n}$ (by the recursive formula)
$=\frac{2n+1}{2(n+1)} \cdot \frac{(2n)!\pi}{2^{2n+1}(n!)^2}$
$=\frac{(2n+1)(2n+2)}{2^2 (n+1)^2} \cdot \frac{(2n)!\pi}{2^{2n+1}(n!)^2}$

It's easy to prove that this equals $\frac{(2(n+1))! \pi}{2^{2(n+1)+1} [(n+1)!]^2}=\frac{(2n+2)!\pi}{2^{2n+3} [(n+1)!]^2}$, which is what you are looking for.

Question 2. > similarly, find a recursive formula for $I_{2n-1}$

5. WOW!, than you so much!, i never once thought to use induction, ive spent near enough 5 hours trying to show it n i jus coudlnt get it