# 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