During a quiz my Calc 3 professor made a typo. He corrected it in class, but he offered a challenge to anyone who could solve the integral.
The (original) question was:
Find the length of the curve as described by the vector valued function $\vec{r} = \frac{1}{3}t^{3}\vec{i} + t^{2}\vec{j} + 4t\vec{k} $ where $0 \le t \le 3$
This give us:
$\int_0^3 \! \sqrt{t^{4}+4t^{2}+16} \, \mathrm{d}t$
Wolfram Alpha says that the solution to this involves Incomplete Elliptic Integrals of the First and Second Kinds. I was wondering if anyone had any level appropriate resources where I can find information about how to attack integrals like this.
Thanks in advance.
Best Answer
(might as well...)
As a quick review: anytime you see an integral that involves the square root of a cubic or a quartic polynomial, it is quite likely that an elliptic integral will be needed (except in some very special cases, termed "pseudoelliptic integrals").
Now, for the integral at hand (treating the indefinite case for now, and worrying about the limits later): your integrand is a quartic that consists of only even powers of the dummy variable, so the first thing to do is to perform a (modified) Weierstrass substitution, $t=2\tan\dfrac{u}{2}$:
$$\begin{align*} \int \sqrt{t^4+4t^2+16} \,\mathrm dt&=8\int \frac{\sqrt{3+\cos^2 u}}{(1+\cos\,u)^2} \mathrm du=8\int \frac{\sqrt{4-\sin^2 u}}{(1+\cos\,u)^2} \mathrm du\\ &=16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du \end{align*}$$
(the additional factor of $2$ in the substitution used comes from the fourth root of the polynomial's constant term.)
Why did I make those last few transformations, you ask? This is because the standard elliptic integrals are made to pop out easily when the part within the square root takes the form $\sqrt{1-m\,\sin^2 u}$.
In any event, we now need to perform a Jacobian substitution. We let $u=\mathrm{am}(v\mid m)$, where $\mathrm{am}$ is the Jacobian amplitude function and $m$ is a constant (in elliptic integral parlance, the parameter) to be determined. The usual Jacobian elliptic functions come from this basic function:
$$\begin{align*} \mathrm{sn}(v\mid m)&=\sin(\mathrm{am}(v\mid m))\\ \mathrm{cn}(v\mid m)&=\cos(\mathrm{am}(v\mid m))\\ \mathrm{dn}(v\mid m)&=\frac{\mathrm d}{\mathrm dv}\mathrm{am}(v\mid m)=\sqrt{1-m\,\sin^2(\mathrm{am}(v\mid m))} \end{align*}$$
Consideration of the last relation involving $\mathrm{dn}$ indicates that we can choose $m=\frac14$ for convenience; thus,
$$16\int \frac{\sqrt{1-\frac14 \sin^2 u}}{(1+\cos\,u)^2} \mathrm du=16\int \frac{\mathrm{dn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv$$
Unfortunately, things get rather messy at this juncture, so I had to ask for some assistance from Mathematica. A little bit of massaging of the results from Mathematica yielded
$$4\int \frac{3+\mathrm{cn}^2\left(v\mid\frac14\right)}{\left(1+\mathrm{cn}\left(v\mid\frac14\right)\right)^2} \mathrm dv=\frac23\left(6v-4\,\varepsilon\left(v \mid \frac14\right)+\frac{4\left(3+\mathrm{cn}\left(v\mid \frac14\right)\right)\mathrm{dn}\left(v\mid \frac14\right)\mathrm{sn}\left(v\mid \frac14\right)}{\left(1+\mathrm{cn}\left(v\mid \frac14\right)\right)^2}\right)$$
where $\varepsilon(v\mid m)$ is the Jacobi epsilon function.
One could of course use the appropriate formulae from Byrd and Friedman's Handbook of Elliptic Integrals for Engineers and Scientists if one insists on a wholly manual evaluation, but the evaluation involves a rather messy set of recurrence relations. (If I find more time, I'll eventually edit this answer to include the full solution, bloody guts and all...)
In any event, to now consider the definite integral, we undo the two substitutions in turn and apply those to the limits of the original integral. In particular, the inverse of $\phi=\mathrm{am}(v\mid m)$ is the incomplete elliptic integral of the first kind, $v=F(\phi\mid m)$, while the Jacobi epsilon function satisfies the relation
$\varepsilon(F(\phi\mid m)\mid m)=E(\phi\mid m)$
where $E(\phi\mid m)$ is the incomplete elliptic integral of the second kind. The transformed limits are then $v=0$ and $v=F\left(2\arctan\frac32\mid\frac14\right)$
We finally end up with
$$\int_0^3 \sqrt{t^4+4t^2+16} \,\mathrm dt=\frac{17}{13}\sqrt{133}+4\,F\left(2\arctan\frac32\mid\frac14\right)-\frac83 \,E\left(2\arctan\frac32\mid\frac14\right)$$
which is a fair bit simpler than what you get if you directly input the integral into Mathematica.
If it wasn't already apparent in the paragraphs above: elliptic integrals and elliptic functions are very different things; one is (very roughly) the inverse of the other. Too many people tend to conflate these two families of functions, though they are definitely related...
In addition to Byrd/Friedman, you'll also want to look into Greenhill's The Applications of Elliptic Functions, which, contrary to the title, also deals a bit with the elliptic integrals as well.