- #1

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- Thread starter jamie516
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- #1

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- #2

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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)^{-1}dx

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

(cos

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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- #3

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yes, they are constants

- #4

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[tex]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[/tex]

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

- #5

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

- #6

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

- #7

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[tex]

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}

[/tex]

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

Regards.

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}

[/tex]

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

Regards.

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- #8

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- #9

MathematicalPhysicist

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(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.

- #10

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- #11

djeitnstine

Gold Member

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Code:

```
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

Code:

```
integrate(sec(atan(a*cos(x))), x);
[tex]\int{\sqrt{a^2cos^2(x)+1}}dx[/tex]
```

- #12

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wolfram mathematica got an answer for the integral:

http://www.workmad3.com/MSP22610029998209164948_87.gif [Broken]

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 :)

http://www.workmad3.com/MSP22610029998209164948_87.gif [Broken]

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 :)

Last edited by a moderator:

- #13

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- #14

djeitnstine

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

- #15

Mute

Homework Helper

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[tex]\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 [/tex]

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

[tex]\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 [/tex]

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.

- #16

djeitnstine

Gold Member

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[tex]\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 [/tex]

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

[tex]\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 [/tex]

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

- #17

Mute

Homework Helper

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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.

- #18

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