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DrKareem
Jan23-07, 07:05 PM
Hi, the integral I'm trying to solve is:

\int \frac{1}{1+\sin^2{\theta}}d\theta

From 0 to pi, which is the same as half the same integral over the unit circle.

I changed the sine squared into 1-cos[2x] and then expressed cos[2x] as half z^2 +(z-1)^2 and i finally got it in this form:

\frac{1}{i} \int \frac{z}{-z^4+z^2-1} dz

I found out the poles to be -1/2 +- i sqrt(3)/2 which happen to be on the countour path, and in that case i don't know how to apply the residue theorem, if it could be applied that is.

Edit: The answer is given in the book to be pi/sqrt(2)

Gib Z
Jan23-07, 07:55 PM
Let sin(x)=u and express the entire integral in terms of u. the denominator is just 1+u^2, and solve and sub to dtheta to du, we get

\int \frac{\sqrt{1-u^2}}{1+u^2} du. Trig sub.

Gib Z
Jan23-07, 08:12 PM
Ahh nvm my last post..I must have done something wrong cuz when you check it on www.calc101.com it dusnt gimme the same thing...But I dont undersatnd how you textbook gives you a constant for an indefinite integral..

BTW: The answer isnt nice..
Your textbook is correct if this is pi:

DrKareem
Jan23-07, 08:12 PM
Okay I see where you're going here, it should be solved using that method. However, i took the integral from a chapter on the residue method in integration. So i'm trying to solve it using that method, and explained how i got stuck.

Gib Z
Jan23-07, 08:19 PM
I have no idea what the theorem is :D

Edit: Method, not theorem

joob
Jan23-07, 08:35 PM
your sub to cos(2x) is fine, and then exp's. what you then want to do is substitute z=e^{2 i \theta} thus dz=2i z d\theta . Your substitution was a little funky. thus cos(2 \theta) = \frac{z+ (1/z)}{2} .

joob
Jan23-07, 08:36 PM
youll get a slightly easier integral which you can just take the residue of for the answer

DrKareem
Jan23-07, 08:42 PM
Oh my God that was so trivial. Thanks for the help guys.

Edit: Actually it IS the residue THEOREM not method :P

thebuttonfreak
Jan23-07, 10:01 PM
cool problem, shows the awesome power of the residue theorem!

DrKareem
Jan24-07, 09:53 AM
Yes awesome indeed. I just started studying this and it looks really nice, although some aspects of it are hard to grasp. I guess i should be practicing even more...

complexPHILOSOPHY
Jan24-07, 12:12 PM
Practice makes permanant!