## zepdrix 2 years ago Calculus 3, Need help with Product Notation.

1. zepdrix

So I have this annoying project I'm working on... I think I've gotten it pretty far. I end up the product of integrals like this. $\int\limits_0^{\pi/2}\cos^2x dx \cdot \int\limits_0^{\pi/2}\cos^4x dx\cdot \int\limits_0^{\pi/2}\cos^6x dx \cdot\cdot\cdot \int\limits_0^{\pi/2}\cos^nx dx$ Which, if I've done my math correctly, simplifies to something like this,$\left[\left(\frac{\pi}{4}\right)\left(\frac{\pi}{4}\cdot\frac{3}{4}\right)\cdot\cdot\cdot\left(\frac{\pi}{4}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{n-1}{n}\right)\right]$

2. zepdrix

Is there a way I can simplify this using the Product notation? I'm just not very familiar with it D:

3. zepdrix

$\huge \frac{\pi}{4}\quad \prod_{i=1}^{n-3}\quad\frac{i+2}{i+3}$Like that maybe? :o bah I dunno what I'm doing.. lol

4. zepdrix

Hmm no I guess that doesn't work :O that gives me,$\left[\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{n-1}{n}\right]$It's not repeating the terms each time. Maybe there's not a nice way to simplify it? :x

5. abb0t

Fubini's theorem?

6. zepdrix

Hmm I don't remember that :O I should look that up. Oh I should have been a little clearer in the first post also. Each of those integrals is a different X. x1, x2 and so on :O

7. KingGeorge

Sorry for very late reply, but you can write it as $\large \left(\frac{\pi}{4}\right)^{n-2}\quad \prod_{i=1}^{n-3}\quad\frac{i+2}{i+3}$The product was correct, you just needed to factor out the $$\pi/4$$ terms. If you don't want to do that, You could have written $\large \frac{\pi}{4}\quad\prod_{i=1}^{n-3}\left[\frac{\pi}{4}\cdot\frac{i+2}{i+3}\right]$