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Did anyone else get <math>5\sin{\theta} + C</math>, a much simpler answer than in the back of the book? | Did anyone else get <math>5\sin{\theta} + C</math>, a much simpler answer than in the back of the book? | ||
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==[[Answers to Even Questions_MA181Fall2008bell]]== | ==[[Answers to Even Questions_MA181Fall2008bell]]== | ||
Revision as of 12:33, 19 October 2008
8.5 #20
So I finally solved this one after a bit of brain failure. In case anyone else gets the same point as me and is wondering how in the world to integrate
$ \int\frac{d\theta}{1+\cos\theta} $
Stop, you've went too far. Go back a few steps until you have something like:
$ \int\frac{d\theta}{\cos^2\theta} $
And integrate from there. If you don't know how, remember your very basic trig identities. Like that 1/cos(x) = sec(x). Yeah, that should help.
Boy did I feel stupid when I finally figured it out, like ten minutes after seeing if I had to integrate by parts and then asking another math nerd on my floor to help (disclaimer: he didn't see it at first either. It wasn't until we sat down and were really gonna start working on it that I was like, "Hey, wait, what is one over cosine squared? Isn't that secant squared?"). So yeah, stupid moment. Hope no one else gets as stuck as I did. But hey, that's what this is for, right?
Pfff, quitter. With a little work, you can show:
$ \int\frac{d\theta}{\cos^2\theta} = 2\int\frac{d\theta}{1+\cos 2\theta} = \csc 2\theta - \cot 2\theta + C $
Which is clearly the better answer. --John Mason 17:26, 17 October 2008 (UTC)
Incidentally, this is a half-angle formula for tangent:
$ \tan\frac{\theta}{2} = \csc \theta - \cot \theta = \frac{1 - \cos \theta}{\sin \theta} $
--John Mason 21:08, 17 October 2008 (UTC)
Page 575, Problem 7
Did anyone else get $ 5\sin{\theta} + C $, a much simpler answer than in the back of the book?
Oh, I guess I have to change it back so it's in terms of t.
