# 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!