A nasty integral to compute

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1. Jan 20, 2016

Rmehtany

Hey Guys!

I was working on an integration problem, and I "simplified" the integral to the following:

$$\int \limits_0^{2\pi} (7.625+.275 \cos(4x))^{1.5} \cdot (A \cos(Nx) + B \sin(Nx)) \cdot (Z-v \cos(x)) dx$$

This integral may seem impossible (I have almost lost all hope on doing this analytically.) If anyone can suggest a approach on attacking this monster, please suggest (It doesn't matter how ugly the solution will turn up to be. I will churn it out.)

Note: Assume all values to be constants EXCEPT $x$

ATTEMPTS:

An interesting idea to solve this was to use the exponential notation to replace the trigonometric functions. An example:

$$\cos(Nx) = \frac{1}{2}(e^{Nix} + e^{-Nix})$$

This allows me to express the equation in terms of these exponent terms, which is nice because the magnitude of $e^{ix}$ is 1, making the integral a closed loop integral $$\oint$$ with the integration variable equal to one. I tried to use reverse Green's theorem, but I got stuck

Suggestions?

2. Jan 20, 2016

mfig

Mathematica seems unable to do it. That is not a good sign for it having a closed form solution. Is there some reason you don't simply evaluate this numerically?

Here is the result of trying.

3. Jan 20, 2016

Rmehtany

This is for my research project, and for reasons that are too complicated to explain now, I need a numerical answer. I have some evidence based on partial testing that the integral has a closed form solution. This includes the form that I discussed in the second part of my first post

4. Jan 20, 2016

mfig

Numerically evaluating the integral will give you that. It should be very easy to numerically integrate this using MATLAB or whatever.

5. Jan 20, 2016

phi1123

Well if you're really interested in getting an analytical solution, you could try expanding the $(7.625+.275 \cos(4x))^{1.5}$ as a taylor series. At that point you'd have a sum of integrals that are just powers of trig functions, which should be integrable (though ugly). You could then truncate the series to get a number; should be pretty accurate since 0.275<<7.625