Elliptic integrals

[BlackHole]

Does anyone have any idea how i can find the antiderivative of cos^3/2x in terms of simple functions?

 

The result is [math]\frac{\sqrt \cos{x}}{1/2}[/math]. Ive been playing with it, cant really get anywhere. I was trying to figure out a way with trig identities to simplify it, but I can't get rid of the square root.

 

Maybe using Jacobi's elliptic functions will solve the problem.

[Johnny5]

Does anyone have any idea how i can find the antiderivative of cos^3/2x in terms of simple functions?

 

The result is [math]\frac{\sqrt \cos{x}}{1/2}[/math]. Ive been playing with it' date=' cant really get anywhere. I was trying to figure out a way with trig identities to simplify it, but I can't get rid of the square root.

 

Maybe using Jacobi's elliptic functions will solve the problem.[/quote']

 

 

[math] \int ( cos x )^{3/2} dx [/math]

 

Let me think about it.

 

[math] \frac{d}{dx} ( cos x )^{1/2} = 1/2 (cos x )^{-1/2} (- sin x )+ C [/math]

 

[math] 2(cos x )^{1/2} \frac{d}{dx} ( cos x )^{1/2} = - sin x + C^\prime [/math]

 

Let

[math] U = ( cos x )^{1/2} [/math]

 

[math] 2U \frac{dU}{dx} = - sin x + C^\prime [/math]

 

[math] U \frac{dU}{dx} + U \frac{dU}{dx} = - sin x + C^\prime [/math]

 

[math] U^2 = cos x [/math]

 

d(UV) = VdU+UdV

 

d(UU)=UdU+UdU

 

[math] \frac{d}{dx} U^2 = U \frac{dU}{dx} +U \frac{dU}{dx} [/math]

 

[math] \frac{d}{dx} U^2 = - sin x + C^\prime [/math]

 

[math] d(U^2) = - sin x dx + C^\prime dx [/math]

 

[math] \int d(U^2) = \int (- sin x dx + C^\prime dx ) [/math]

 

[math] U^2 + C_1 = \int - sin x dx + C^\prime \int dx [/math]

 

[math] (cos x)^2 + C_1 = \int - sin x dx + C^\prime \int dx [/math]

 

[math] (cos x)^2 + C_1 = - \int sin x dx + C^\prime (x+c_2) [/math]

 

[math] (cos x)^2 + C_1 = - (-cos x + C_3) + C^\prime (x+c_2) [/math]

 

[math] (cos x)^2 + C_1 = cos x - C_3 + C^\prime (x+c_2) [/math]

 

[math] (cos x)^2 + C_1 = cos x - C_3 + C^\prime x+ C^\prime c_2 [/math]

 

[math] (cos x)^2 = cos x - C_3 + C^\prime x+ C^\prime c_2 - C_1[/math]

 

[math] (cos x)^2 = cos x + C^\prime x - C_3 + C^\prime c_2 - C_1[/math]

 

Let

 

[math] D = - C_3 + C^\prime c_2 - C_1 [/math]

 

[math] (cos x)^2 = cos x + C^\prime x + D [/math]

 

[math] (cos x)^2 - D - C^\prime x = cos x [/math]

 

Let

 

[math] v^2 = D + C^\prime x [/math]

 

[math] (cos x)^2 - v^2 = cos x [/math]

 

[math] (cos x + v )(cos x - v)= cos x [/math]

 

Consider a case where v=0. If such a case is possible then:

 

[math] (cos x)^2= cos x [/math]

 

In such a case we have:

 

[math] 0 = D + C^\prime x [/math]

 

From which it follows that:

 

[math] C^\prime x = -D = C_3 - C^\prime c_2 + C_1[/math]

 

From which it follows that: If not (C` = 0) then

 

[math] x = C_3/C` - c_2 + C_1/C`[/math]

 

So in such a case, x is a constant, hence dx=0.

 

Recall, we are trying to solve this:

 

[math] \int ( cos x )^{3/2} dx [/math]

 

The answer is an unknown function of X, which we can denote by f(x), plus an arbitrary constant of integration which I will denote by A. So in other words:

 

[math] \int ( cos x )^{3/2} dx = f(x) + A [/math]

 

Now, in the case where dx=0 we have this:

 

[math] \int ( cos x )^{3/2} 0 = 0 = f(x) + A [/math]

 

In which case, f(x) = -A

 

You have stated that the answer is given by:

 

[math] f(x) = 2 \sqrt{cos x} [/math]

 

I assumed that is the answer, so that means this:

 

[math] 2 \sqrt{cos x} = -A [/math]

[BlackHole]

From integration by parts i got:

 

[math]\cos^{3/2}x = \cos x (cos^{1/2})[/math]

 

[math]u = \cos^{1/2}x ; dv = \cos x ; v = \sin x[/math]

 

[math]\cos^{1/2}x (\sin x - {1/2}) \int \sin x^{3/2}x dx[/math]

 

That's a problem in itself...

 

I think we need to use [math]F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}[/math].

[Meir Achuz]

I am puzzled by your question. As Johnny5's first step shows

(The constant C should not be there.), the answer you give is wrong.

[Dave]

Plugging it into mathematica would suggest that elliptic integrals are indeed needed. However, I don't really have any experience at all in this field. If anyone else cares to step in...?

[BlackHole]

Elliptic functions are beyond me. I don't know whether that's required for QM and especially GR though.

 

Edit: An EllipticF is the eliptic integral of the first (F) kind defined as:

 

[math]F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}[/math].

 

Mathematica gave this answer:

 

[math]\int \cos^{3/2}x \ dx = \frac{2}{3} (EllipticF[\frac{x}{2}, 2]) + \sqrt {cos x} \ sin x[/math]

 

Elliptic functions are not widely used in physics (except maybe dynamics) but anyway.

[Johnny5]

What I was doing the other day didn't go anywhere. In fact, I shouldn't have approached the problem the way I did. You gave part of the answer, but I shouldn't have tried to go from the answer you gave to the question.

 

I happen to have my own arsenal for solving integrals. Let me have a look at this guy...

 

[math]F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}[/math].

[BlackHole]

What I was doing the other day didn't go anywhere. In fact' date=' I shouldn't have approached the problem the way I did. You gave part of the answer, but I shouldn't have tried to go from the answer you gave to the question.

 

I happen to have my own arsenal for solving integrals. Let me have a look at this guy...

 

[math']F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}[/math].

 

No, i believe that what you did was correct. The integration by parts was correct but first we need to find the antiderivative because elliptic functions are inverse to elliptic integrals.

 

Jacobi's elliptic functions (which are the closest to elliptic integrals) arise as solutions to the form:

 

[math] \frac{d^{2x}}{dt^2} = A + Bx + Cx^2 + Dx^3 [/math]

[Meir Achuz]

An elliptic integral is an integral. Converting one integral to another integral is another integral, and not an answer.