anonymous
  • anonymous
integrate sin3theta*sin5theta dtheta. I've done the problem a few times but keep coming up with the wrong answer.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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zepdrix
  • zepdrix
What methods have you tried? Hmm I think `by parts` will work very nicely.
anonymous
  • anonymous
I was using 1/2(cos(m-n)x-cos(m+n)x)
zepdrix
  • zepdrix
Hmm I don't know what that is D:

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anonymous
  • anonymous
it's a set of three formulas we got in class. It was probably derived from integration by parts though originally
anonymous
  • anonymous
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zepdrix
  • zepdrix
grrr you should have rotated it :) lol
anonymous
  • anonymous
haha, sorry
zepdrix
  • zepdrix
Hmm that's a nifty little formula. So I guess that gives us,\[\large \int\limits \sin5\theta \sin3\theta\;d \theta \quad=\quad \frac{1}{2}\int\limits \cos2\theta-\cos8\theta\;d \theta\]right?
anonymous
  • anonymous
I did a -2theta for m-n
zepdrix
  • zepdrix
So you ended up with,\[\large \int\limits\limits \sin5\theta \sin3\theta\;d \theta \quad=\quad \frac{1}{2}\int\limits\limits \cos(-2\theta)-\cos8\theta\;d \theta\]
anonymous
  • anonymous
yes
zepdrix
  • zepdrix
Keep in mind that cosine is an `even function` so:\[\Large \cos(-2\theta) \quad=\quad \cos(2\theta)\]
anonymous
  • anonymous
ah, okay, cool
zepdrix
  • zepdrix
So what did you do from that step? Did you do a `u-substition` ? You'll want to try and get more comfortable with problems like this, so you don't have to bother with a u-sub. On integrals like this, we simply divide by the coefficient on theta when we integrate.
zepdrix
  • zepdrix
Example:\[\Large \int\limits \cos(4\theta)\;d \theta \quad=\quad \frac{1}{4}\sin(4\theta)\] Understand what I mean? :o We just divide by the coefficient (the 4), instead of doing a fancy u-sub. If that's too confusing you can go ahead with a u-sub though :)
anonymous
  • anonymous
that's what I was doing. The signs probably messed me up
anonymous
  • anonymous
yeah, definitely. I don't mess with u subs on those simple ones
zepdrix
  • zepdrix
\[\large \frac{1}{2}\int\limits\limits \cos2\theta-\cos8\theta\;d \theta \quad=\quad \frac{1}{2}\left(-\frac{1}{2}\sin(-2\theta)-\frac{1}{8}\sin8\theta\right)\]
zepdrix
  • zepdrix
So you end up with something like that?
zepdrix
  • zepdrix
Woops I meant to put cos(-2theta) in the original, to match what you've been working with. Same result though.
anonymous
  • anonymous
that is what I've been getting
zepdrix
  • zepdrix
So why do you think your answer is wrong? :3 maybe it just looks a little different than the book.
anonymous
  • anonymous
:/
zepdrix
  • zepdrix
From here, you would want to remember that sine is an `odd function`:\[\Large \sin(-2\theta) \quad=\quad -\sin(2\theta)\]
zepdrix
  • zepdrix
So that's why that negative ends up disappearing on the first term.
anonymous
  • anonymous
We do our homework online and it keeps counting my answer wrong. Online systems are so picky though. Glad I'm doing it right though
zepdrix
  • zepdrix
Oh yah, that can be a bit tricky D:
zepdrix
  • zepdrix
A bit frustrating i mean*
anonymous
  • anonymous
Thank you for your help though. And btw, I've mastered that reduction formula now :P
zepdrix
  • zepdrix
Oh cool c:

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