# Quite a complex integral

• jamie516

#### jamie516

Ok, I am trying to integrate the following function, and not getting very far: it's s=integral between 0 and 2pi of cos^-1(arctan((2*pi/b)*a*cos(2*pi*x/b)))dx)^-1 where a and b are known variables. What I would like to know, is can this integral be evaluated directly, or must I use the trapezium rule, and if so how would I do that?

by known variables, do you mean constant in the formula? If so, it should be possible to integrate as what you then have is
(cos-1(X*cos(2pi * x/b)-1dx
to integrate. You should be able to integrate that although I can't remember what the integral of the inverse trig functions are off the top of my head. It may be a pretty complex integral but I'd expect it to be possible to do so with a closed-form solution.

edit: Ignore that... I misread the original formula and didn't match up the brackets... need to re-evaluate

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yes, they are constants

Just trying to visualize the formula.

$$s= \displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx )^-1$$

What about the last missing ")^-1"?

That is correct, and the whole integral is ^-1.

so you would need a ( in front of the integral sign to complete it

$$s= \dfrac{1}{\displaystyle\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx}$$

I couldn't solve it. Neither did "Derive 6".

Regards.

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yes, that is right, and a and b can be any numbers really, as they define the shape of the sine wave.

I haven't done any calculuations yet, but one thought that got into my attention, is perhaps you should take derivative under the integral sign of a or b, and use the theorem, that:
(d/da)S f(x,a)dx=S df(x,a)/da dx.
Not sure if that will ease the calculations, but it sure as hell worth the effort.

Cheers.

Maxima got this from your basic function without the constants

Code:
integrate(sec(atan(cos(x))), x);

$$\int{\sqrt{cos^2(x)+1}}dx$$

which of course has no analytical solution =(

edit: with a constant such as the 'a' here

Code:
integrate(sec(atan(a*cos(x))), x);

$$\int{\sqrt{a^2cos^2(x)+1}}dx$$

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This is a DEFINITE integral which means I'm sure that a computer program can make a numeric approximation, hopefully.

This is a DEFINITE integral which means I'm sure that a computer program can make a numeric approximation, hopefully.

certainly approximations can be made but the OP would like to have an analytical answer.

Waaaaaaiiiiiit just a minute here. The integral that GRfrones suggested,

$$\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx$$

does not appear to be the integral in the original post (following the standard convention of notations):

"integral between 0 and 2pi of cos^-1(arctan((2*pi/b)*a*cos(2*pi*x/b)))dx"

GRfrones seems to have interpreted "cos^(-1)(stuff)" as "sec(stuff)", but I would take this to be "arccos(stuff)", as that's pretty much exclusively what "cos^(-1)(stuff)" means, and so I would have thought the integral is

$$\int_0^{2\pi} \cos^{-1}\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx$$

Is it really GRfrones' interpretation? If so, then the OP should take care when writing "cos^(-1)" to mean "sec", as I can think of no instance where "cos^(-1)" would be taken to mean the secant.

Waaaaaaiiiiiit just a minute here. The integral that GRfrones suggested,

$$\int_0^{2\pi} \sec\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx$$

does not appear to be the integral in the original post (following the standard convention of notations):

"integral between 0 and 2pi of cos^-1(arctan((2*pi/b)*a*cos(2*pi*x/b)))dx"

GRfrones seems to have interpreted "cos^(-1)(stuff)" as "sec(stuff)", but I would take this to be "arccos(stuff)", as that's pretty much exclusively what "cos^(-1)(stuff)" means, and so I would have thought the integral is

$$\int_0^{2\pi} \cos^{-1}\left(\arctan\left(\left(\dfrac{2\pi}{b}\right ) a \cos\left(\dfrac{2\pi x}{b}\right)\right)\right)dx$$

Is it really GRfrones' interpretation? If so, then the OP should take care when writing "cos^(-1)" to mean "sec", as I can think of no instance where "cos^(-1)" would be taken to mean the secant.

The OP has already identified the correct integral

The OP has already identified the correct integral

I'm aware of that, I wanted to double-check, and point out that if that is the correct integral then the use of "cos^{-1}" to mean "sec" is not a good idea, as it conflicts with the standard usage of the symbol.

Changing it from sec to arccos stops the integral from working on wolfram as well. I'm not sure that is a good or bad thing though, considering the integral it returns for the one identified as correct by the OP :)