Help with an integral, yet again.

1. Jan 23, 2007

DrKareem

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)

Last edited: Jan 23, 2007
2. Jan 23, 2007

Gib Z

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.

Last edited: Jan 23, 2007
3. Jan 23, 2007

Gib Z

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..

Your textbook is correct if this is pi:

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Last edited: Jan 23, 2007
4. Jan 23, 2007

DrKareem

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.

5. Jan 23, 2007

Gib Z

I have no idea what the theorem is :D

Edit: Method, not theorem

6. Jan 23, 2007

joob

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 $$cos(2 \theta) = \frac{z+ (1/z)}{2}$$.

7. Jan 23, 2007

joob

youll get a slightly easier integral which you can just take the residue of for the answer

8. Jan 23, 2007

DrKareem

Oh my God that was so trivial. Thanks for the help guys.

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

9. Jan 23, 2007

thebuttonfreak

cool problem, shows the awesome power of the residue theorem!

10. Jan 24, 2007

DrKareem

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...

11. Jan 24, 2007

complexPHILOSOPHY

Practice makes permanant!