Difference between revisions of "Homework 13 MA181Fall2008bell" - Rhea

Revision as of 10:54, 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)

Answers to Even Questions_MA181Fall2008bell