# Integral of sqrt(x^3 -1) dx

integral of sqrt(x^3 -1) dx.
I have tries it with trigonometri subs, but I only found till
integral of 2/3(sin^5/3 (tetha) / cos^2 (tetha) ) d(tetha)

I don't believe this has a simple antiderivative.

hm you may be right

Yes, this requires the use of elliptical integrals (non-elementary functions).

what's that? I don't know about it...

mathwonk
Homework Helper
that is sort of a tautological stetament, i.e. elliptic functions are defined in terms of integrals, of (usually reciprocals of) square roots of cubics.

i.e. just as sin can be defiend as the inmverse of the integral of dx/sqrt(1-x^2), so also one can define some interesting fucntions as the inverse of the integral of

things like dx/sqrt(1-x^3).

wow... can someone walk me through how this would work out

I just typed it in and got

$$\int \sqrt{x^3-1}dx=\frac{2x\sqrt{x^3-1}}{5}-\frac{3\int{\frac{1}{\sqrt{x^3-1}}}}{5}$$

dextercioby
Homework Helper
Okay,here's the result for the curious.

Daniel.

#### Attachments

• 2.3 KB Views: 4,802
ummm I don't really follow what ur attachment is showing.

dextercioby
Homework Helper
You mean the elliptic integral...?Go to wolfram's site (mathworld) and search for Legendre elliptic integrals.

Daniel.

saltydog
Homework Helper
The formula Daniel reported is the anti-derivative. That is:

$$\int\frac{1}{\sqrt{x^3-1}}=\frac{2i\sqrt{(-1)^{5/6}(x-1)}\sqrt{1+x+x^2}EllipticF[\arcsin(\frac{\sqrt{-(-1)^{5/6}-ix}}{3^{1/4}}),(-1)^{\frac{1}{3}}]}{3^{\frac{1}{3}}\sqrt{x^3-1}}$$

It's just not in terms of elementary functions.

Great. Now suppose I want to integrate it from 2 to 4 using the formula and for the moment I want to concentrate on the elliptical integral portion, the EllipticF part. A definition first. EllipticF is the eliptical integral of the first (F) kind defined as:

$$F(x,k)=\int_0^x\frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}}$$

Alright, it's already getting messy. Let's just concentrate on the ArcSin part first then: Let x=2 so we have:

$$\arcsin[\frac{\sqrt{-(-1)^{5/6}-2i}}{3^{1/4}}]$$

Jesus. That's a problem in itself (for me anyway). Will need to spend time on it first. Anyway, I want to see how the imaginary parts of the anti-derivative are eliminated since this is a definite integral. I've attached a plot of the function.

#### Attachments

• 3.4 KB Views: 953
Last edited:
saltydog
Homework Helper
Alright, I'm stuck. Let's just concentrate on the radical:

$$\sqrt{-(-1)^{5/6}-2i}$$

Sinking deeper still . . . how about just the -1 to the 1/6 power:

$$\sqrt[6]{-1}$$

Wouldn't I get six roots from this? Which one do I use? Going to bed . . . will work on it latter. Any comments would be appreciated though.

that is sort of a tautological stetament, i.e. elliptic functions are defined in terms of integrals, of (usually reciprocals of) square roots of cubics.
If we define one new function, and it allows to solve a whole new class of integrals, are these solutions tautological?

Answer: Yes, but that is the way math is. Mathematical satisfaction comes from getting so used to the tautologies that they talk on a life of there own:

Von Neumann: "You don't understandthings in mathematics, you just get used to them".

saltydog
Homework Helper
Well, I'm still working on the elliptical integral expression and finding it interesting. You know, even the part:

$$EllipticF[\arcsin(\frac{\sqrt{-(-1)^{5/6}-ix}}{3^{1/4}}),(-1)^{\frac{1}{3}}]$$

is a challenge for me (well, I don't mean just plug it into Mathematica and turn the crank you know but I am using it to study the expression). Seems the definite integral ends up being (a+bi)-(c+bi) for any limits! That's how the i is eliminated. How's that? I'll work with it more.

I got this problem from someone by forum too. I thought it wasn't so difficult.. thanx for reply. but is the function really can be integrated?

dextercioby