Contour Integrals

• May 6th 2010, 11:17 AM
Maths123
Contour Integrals
How do I solve the following:

I= Integral between 0 and 2pi [4dx/(2cosx+ 3sinx + 2i)]

Thanks
• May 7th 2010, 02:24 AM
piglet
Quote:

Originally Posted by Maths123
How do I solve the following:

I= Integral between 0 and 2pi [4dx/(2cosx+ 3sinx + 2i)]

Thanks

well what i'd normally do is use the following substitutions

let $cosx = \frac{1}{2}(x + \frac{1}{x})$ and $sinx = \frac{1}{2i}(x+\frac{1}{x})$

If you have any trouble from there on, just ask
• May 11th 2010, 11:34 PM
Maths123
Right, I subbed that in.

It's not working for me. I don't understand how you then go on to evaluate it further.

Thanks
• May 11th 2010, 11:49 PM
Maths123
I took 4 out, so got

1/(x+(1/x)+(3/2i)x+ (3/2ix)+2i)

Where do I go from there?
• May 12th 2010, 06:23 AM
Maths123
Ok, so I get (2±(17)^0.5)/2i+3

Is that what you get? Where do I go from there?

Thanks
• May 23rd 2010, 05:34 AM
GeoC
Is this integral to be evaluated on a closed path about the origin of the complex plane, or is this a line integral along the real x-axis?