I need to calculate the definite integral of dx/(4+3cos(x)) from 0 to 2pi.(adsbygoogle = window.adsbygoogle || []).push({});

I believe that the integral from 0 to 2pi of (f(e^ix))*i*(e^ix)dx is equivalent to the integral over the unit circle of f(z)dz.

If that's true, then this problem boils down to finding f(z) so that f(z)iz=1/(4+(3/2)(z+z^-1) since cosx=(1/2)(e^ix+e^-1x).

This gives f(z)=-2i/(3z^2+8z+3) which has poles at (-4-√7)/3 and (-4+√7)/3, and only the latter lies within the unit circle, so the value of our integral ought to just be 2*pi*i*Res(f,(-4+√7)/3)=(2*pi*i)*1/((-4+√7)/3-(-4-√7)/3)=2*pi*i*(3/2√7)=6*pi/√7.

However, when I checked my result on Wolframalpha, I was off by a factor of 3. They had 2*pi/√7.

Where did I go wrong?

Note: I also tried to calculate thedefinite integral of dx/(5+4sin(x)) from 0 to 2pi and ended up off by a factor of 2 according again to Wolfram.

Please help.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Using Residue Calculus to evaluate real integrals

**Physics Forums | Science Articles, Homework Help, Discussion**