Thread: Contour Integrals

How do I solve the following:

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

Thanks

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 and

If you have any trouble from there on, just ask

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

I took 4 out, so got

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

Where do I go from there?

Ok, so I get (2±(17)^0.5)/2i+3

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

Thanks

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?