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zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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,\[\left[\left(\frac{\pi}{4}\right)\left(\frac{\pi}{4}\cdot\frac{3}{4}\right)\cdot\cdot\cdot\left(\frac{\pi}{4}\cdot\frac{3}{4}\cdot\cdot\cdot\frac{n1}{n}\right)\right]\]
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Is there a way I can simplify this using the Product notation? I'm just not very familiar with it D:
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
\[\huge \frac{\pi}{4}\quad \prod_{i=1}^{n3}\quad\frac{i+2}{i+3}\]Like that maybe? :o bah I dunno what I'm doing.. lol
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
Hmm no I guess that doesn't work :O that gives me,\[\left[\frac{3}{4}\cdot\frac{5}{6}\cdot\cdot\cdot\frac{n1}{n}\right]\]It's not repeating the terms each time. Maybe there's not a nice way to simplify it? :x
 one year ago

abb0t Group TitleBest ResponseYou've already chosen the best response.0
Fubini's theorem?
 one year ago

zepdrix Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

KingGeorge Group TitleBest ResponseYou've already chosen the best response.0
Sorry for very late reply, but you can write it as \[\large \left(\frac{\pi}{4}\right)^{n2}\quad \prod_{i=1}^{n3}\quad\frac{i+2}{i+3}\]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 \[\large \frac{\pi}{4}\quad\prod_{i=1}^{n3}\left[\frac{\pi}{4}\cdot\frac{i+2}{i+3}\right]\]
 one year ago
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