How do I solve the following:

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

Thanks

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- May 6th 2010, 12:17 PMMaths123Contour Integrals
How do I solve the following:

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

Thanks - May 7th 2010, 03:24 AMpiglet
- May 12th 2010, 12:34 AMMaths123
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 12th 2010, 12:49 AMMaths123
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, 07:23 AMMaths123
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, 06:34 AMGeoC
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?