Can the integral of cos^-1(arctan) be evaluated directly?

[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?

 

[workmad3]

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

 

[jamie516]

yes, they are constants

 

[GRFrones]

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"?

 

[jamie516]

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

 

[jamie516]

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

 

[GRFrones]

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

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

Regards.

 

[jamie516]

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

 

[MathematicalPhysicist]

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.

 

[m00npirate]

Perhaps this could shed some light? It doesn't look pretty though: http://integrals.wolfram.com/index.jsp?expr=sec(arctan((2*pi/b)a*cos(2*pi*x/b)&random=false

 

[djeitnstine]

Maxima got this from your basic function without the constants

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

[tex]\int{\sqrt{cos^2(x)+1}}dx[/tex]

which of course has no analytical solution =(

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

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

[tex]\int{\sqrt{a^2cos^2(x)+1}}dx[/tex]

 

[workmad3]

wolfram mathematica got an answer for the integral:
http://www.workmad3.com/MSP22610029998209164948_87.gif

with the original URL for the equation in case the image stops working:
http://integrals.wolfram.com/index.jsp?expr=sec[arctan[2(pi/b)*a*cos[2pi*x/b]]]&random=false

Edit: It took the program 7.61 seconds to calculate, and the E(x|m) is an 'elliptic integral of the second kind'. Hopefully that means something to someone :)

 

[Pinu7]

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

 

[djeitnstine]

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.

 

[Mute]

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.

 

[djeitnstine]

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

 

[Mute]

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.

 

[workmad3]

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 :)