Does anybody know how to go about solving this integral:
cos(2x+1)cos(4x-2)sin(x) dx
I've tried a number of methods but can't seem to finish it...
You can turn the product $\displaystyle \cos\alpha\cos\beta$ into a sum of cosines, then expand and you'll get terms like $\displaystyle \cos\alpha\sin\gamma$ and then you can turn that product into a sum.
See here: List of trigonometric identities - Wikipedia, the free encyclopedia
I tried that method, or at least, tried to do that method: I ended up with ∫cos(2x+1)cos(2-4x)sinxdx
1) I expanded the cos's that are multiplied by each other with the cosAcosB property.
2) I then used the cosA +cosB property, which simplified to what I wrote above, which isn't really an improvement.
What am I doing wrong?
(Steps would be appreciated)
Not sure how to solve this integral problem? - Yahoo! Answers
Look at the bottom submission, very helpful!
Yes. Very helpful. No doubt you will duly reference this solution in the work you submit (since most colleges, http://berkeley.edu/ for example, otherwise take a dim view of this sort of thing ....)
@ Mr. Fantastic,
We have to reference that? I realize you think it would be plagiarism to take that work and submit it as homework, but that would only be true if one were to completely take someone else's work. I'd already solved most of the problem before I got an answer on there. Just because someone else actually submitted a full length response doesn't mean that I, or anyone else, hadn't solved it before hand. If someone gives you a helpful hint, I view that as tutoring. My school expects you to go get tutoring, but it's not expected that you cite your tutor for their guidance.