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? His Awesomeness, Josh Hunsberger
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.
Answers to Even Questions_MA181Fall2008bell
I am having trouble with #7 on page 569. How do you deal with the fact that sin is to the fourth power? I tried doing integration by parts and that doesn't seem to work. Then I tried making it equal to (1-cos^2)^2, but I don't know where to go from here.
--Klosekam 21:22, 20 October 2008 (UTC)
- Okay, so soemthing went terribly wrong with my browser as i was typing up how to do this and I need to go to bed. I can't show you the fancy version, but I can give you a hint. I'm sure you could do it other ways, but the way I found easiest was to substitute (1-cos2x)/2 for sinx^2 instead of 1-cosx^2. You can then multiply that out. Then you'll have to do it again for the cos^2 term. Finally, when you do that you'll get four very easy integrals, two of which you can add together and two of which will have 0 value for this particular interval. I'm really sorry I couldn't give you a more detailed, prettier explanation, but I'm kinda constrained on time. If you still have trouble, tell me and I'll check here tomorrow around noon and give you a better explanation. Or someone else will.
His Awesomeness, Josh Hunsberger
As a general rule, you substitute with
$ \sin^2 x = 1 - cos^2 x $
if you are dealing with sine to an odd power, and
$ \sin^2 x = \frac{1-\cos 2x}{2} $
if it is an even power. You can think this through by realizing that substituting the former into an even power would just give you the same problem, except with cosines.
