Complex Integral I just can't figure out

Click For Summary
SUMMARY

The integral discussed is ∫ sin²(x)/(3-2cos(x)) dx from x=0 to x=2π. The user attempted to solve it using the substitution z=e^(iθ), leading to the integral i ∫ (z⁴ + 2z² + 1)/(z² - 3z + 1) dz over |z|=1. The user identified poles at z=0 and (3-√5)/2, calculating residues but ultimately found discrepancies with a calculator result of approximately 1.19. The error was traced back to an incorrect substitution for sine, highlighting the importance of accuracy in complex analysis.

PREREQUISITES
  • Complex analysis fundamentals
  • Residue theorem application
  • Complex function substitution techniques
  • Understanding of contour integration
NEXT STEPS
  • Study the residue theorem in complex analysis
  • Practice contour integration with various functions
  • Learn about common substitution techniques in complex integrals
  • Review error-checking methods in mathematical calculations
USEFUL FOR

Students and professionals in mathematics, particularly those focusing on complex analysis, as well as anyone looking to improve their problem-solving skills in integral calculus.

~Death~
Messages
45
Reaction score
0
hi, this is really frustrating because I've been working on this one integral for the last 2 1/2 horus and i can't figure out what I did wrong...

it's the Integral sin^2(x)/(3-2cos(x)) dx from x=0 to x=2pi


I tried the substitution z=e^(itheta) plugging in sin and cos as functions of z and I

ultimately get:

i * integral (z^4+2z^2+1)/(z^2-3z+1)dz over |z|=1

Then the poles inside the unit circle are @ z=0 and (3-sqrt(5))/2

so I find the residue @ z=0 to be 3 and @ (3-sqrt5)/2 something like

(looking in my mess of notes) 4(15-6sqrt(5))^2/((3-sqrt(5)^2(-sqrt(5))

then I plug in 2pi X Sum residues X i

but it doesn't agree with the answer I get on my calculator which is around 1.19

any help is really appreciated thanks!
 
Physics news on Phys.org
sorry to anyone who was trying to figure it out...i solved it

turns out i substitutes 1/2(z+1/z) for sine instead of 1/(2i)(z-1/z)

I make the stupidest mistakes ever and it costs me. anyone know how i can

avoid these =(
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
1K
  • · Replies 27 ·
Replies
27
Views
3K
Replies
2
Views
1K
Replies
4
Views
2K