How do I solve the following: I= Integral between 0 and 2pi [4dx/(2cosx+ 3sinx + 2i)] Thanks
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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 $\displaystyle cosx = \frac{1}{2}(x + \frac{1}{x}) $ and $\displaystyle sinx = \frac{1}{2i}(x+\frac{1}{x}) $ 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?
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