\documentclass[11pt]{article}
\input{/etc/std.tex}
\title{$\int \sqrt{1+t^4}\, dt$}
\author{Brian Bi}
\date{February 12, 2011}
\begin{document}
\maketitle
\section{Introduction}
We wish to evaluate the indefinite integral:
\begin{math}
\int \sqrt{1+t^4} \, dt
\end{math}

Mathematica returns the result:
\begin{equation}
\int \sqrt{1+t^4} \, dt = \frac{t^5 - 2\sqrt[4]{-1} \sqrt{t^4+1}\,
F(i\sinh^{-1}(\sqrt[4]{-1}\, t)\big|-1)+t}{3\sqrt{t^4+1}}+C
\end{equation}
The antiderivative of $f(t) = \sqrt{1+t^4}$ cannot be expressed in terms of
elementary functions. However, Mathematica gives the result in terms of the
special function $F$, a nonelementary function known as the \emph{incomplete
elliptic integral of the first kind}. In the next section this function and its
cousin, the \emph{incomplete elliptic integral of the second kind}, are
introduced. In the following sections we transform the integrand step by step
into a more tractable form.
\section{The $E$ and $F$ functions}
The elliptic integral functions are so named because the context in which they
originally arose was that of finding the arc length of an ellipse. Consider an
ellipse with semimajor axis of length $a$ oriented along the x-axis and
semiminor axis of length 1 oriented along the y-axis. This ellipse has the
parametric equations:
\begin{align}
x &= a \cos t \\
y &= \sin t
\end{align}
(That is, the ellipse is an elongated circle.) As we know, the arc length of the
part of this ellipse defined by $0 \leq t \leq \theta$ is given by
\begin{align}
s(\theta,a) &= \int_0^\theta \sqrt{\left(\frac{dx}{dt}\right)^2+
\left(\frac{dy}{dt}\right)^2}\, dt \\
&= \int_0^\theta \sqrt{a^2 \sin^2 t + \cos^2 t}\, dt \\
&= \int_0^\theta \sqrt{(a^2-1)\sin^2 t + \sin^2 t + \cos^2 t}\, dt \\
&= \int_0^\theta \sqrt{1 + (a^2-1)\sin^2 t}\, dt
\end{align}
Finally, define
\begin{equation}
E(x|k) = \int_0^x \sqrt{1-k \sin^2 t} \, dt
\end{equation}
Then we can write $s(\theta,a) = E(\theta|1-a^2)$, thus expressing the arc
length of the ellipse in terms of this function $E$, which is known as the
\emph{incomplete elliptic integral of the second kind}. It is not possible to
express the function $E$ in terms of elementary functions (exponential, log,
trig functions) of its arguments; if asked to integrate the function
$f(x) = \sqrt{1-k \sin^2 x}$, you have no choice but to reply $E(x|k) + C$.

In some cases we may similarly wish to integrate the function $g(x) =
\frac{1}{\sqrt{1-k\sin^2 x}}$. This function \emph{also} has no elementary
antiderivative; so again mathematicians have given a special name to its
antiderivative, the \emph{incomplete elliptic integral of the first kind}, $F$.
In particular, $F(x|k) + C$ is the antiderivative of $g(x)$.
\section{A hyperbolic substitution}
We wish to integrate $\sqrt{1+t^4}$. It is not at all obvious that the result
can be expressed in terms of incomplete elliptic integrals; we shall have to
proceed through several steps first. The first step is to make the substitution
$t = (-1)^{1/4}\sinh u$, for which $dt = (-1)^{1/4}\cosh u \, du$.
So
\begin{align}
\int \sqrt{1+t^4}\, dt &= \int \sqrt{1 + ((-1)^{1/4}\sinh u)^4}
\,(-1)^{1/4}\cosh u \, du \\
&= (-1)^{1/4} \int \cosh u \sqrt{1 - \sinh^4 u} \, du \\
&= (-1)^{1/4} \int \cosh u \sqrt{1 + \sinh^2 u} \sqrt{1 - \sinh^2 u} \, du \\
&= (-1)^{1/4} \int \cosh^2 u \sqrt{1 - \sinh^2 u} \, du
\label{eqn:orig}
\end{align}
\section{Recast in terms of circular trig functions}
By the definition of the cosh function, and de Moivre's equation:
\begin{align}
\cosh(ix) &= \frac{e^{ix} + e^{-ix}}{2} \\
&= \frac{\cos x + i \sin x + \cos(-x) + i \sin (-x)}{2} \\
&= \frac{2\cos x}{2} \\
&= \cos x
\end{align}
Likewise
\begin{align}
\sinh(ix) &= \frac{e^{ix} - e^{-ix}}{2} \\
&= \frac{\cos x + i \sin x - \cos(-x) - i \sin (-x)}{2} \\
&= \frac{2i\sin x}{2} \\
&= i \sin x \label{eqn:sinh}
\end{align}
To proceed, we employ the substitution $u = ix$, for which $du = i \, dx$. We
then obtain
\begin{align}
\int \cosh^2 u \sqrt{1-\sinh^2 u} \, du
&= \int \cosh^2 (ix) \sqrt{1-\sinh^2(ix)}\, i \, dx \\
&= i \int \cos^2 x \sqrt{1-(i\sin x)^2} \, dx \\
&= i \int \cos^2 x \sqrt{1+\sin^2 x} \, dx
\label{eqn:utox}
\end{align}
\section{Parts}
Now, use the identity $\cos^2 x = \frac{1}{2}(1 + \cos 2x)$:
\begin{align}
J &= \int \cos^2 x \sqrt{1+\sin^2 x} \, dx \label{eqn:J} \\
&= \int \frac{1}{2}(1 + \cos 2x) \sqrt{1+\sin^2 x} \, dx \\
&= \frac{1}{2}\int \cos 2x \sqrt{1+\sin^2 x} \, dx +
\frac{1}{2} \int \sqrt{1+\sin^2 x} \, dx \\
&= \frac{1}{4}\int 2\cos 2x \sqrt{1+\sin^2 x} \, dx + \frac{1}{2}E(x|{-1})
\label{eqn:ident1}
\end{align}
Note that we have used the function $E$; keep watching and this will later
cancel out of the answer, although $F$ will shortly enter the picture.

We tackle the first term by parts:
\begin{align}
\int \sqrt{1+\sin^2 x}\,(2 \cos 2x \, dx)
&= \sqrt{1+\sin^2 x}\int 2 \cos 2x \, dx - \int \left(\int 2 \cos 2x \,
dx\right) \left(\frac{d}{dx} \sqrt{1+\sin^2 x}\right) \, dx \\
&= \sin 2x \sqrt{1+\sin^2 x} - \int \sin 2x \frac{\sin 2x}{2\sqrt{1+\sin^2 x}}
\, dx \\
&= \sin 2x \sqrt{1+\sin^2 x} - \int \frac{\sin^2 2x}{2\sqrt{1+\sin^2 x}}\, dx \\
\label{eqn:parts}
\end{align}
Now
\begin{align}
\int \frac{\sin^2 2x}{2\sqrt{1+\sin^2 x}}\, dx
&= \int \frac{(2\sin x \cos x)^2}{2\sqrt{1+\sin^2 x}}\, dx \\
&= \int \frac{2\sin^2 x \cos^2 x}{\sqrt{1+\sin^2 x}}\, dx \\
&= \int \frac{2\sin^2 x (1-\sin^2 x)}{\sqrt{1+\sin^2 x}}\, dx \\
&= \int \frac{2\sin^2 x - 2\sin^4 x}{\sqrt{1+\sin^2 x}}\, dx \label{eqn:ident2}
\end{align}
Putting it all together we have so far, from (\ref{eqn:ident1}),
(\ref{eqn:parts}), and (\ref{eqn:ident2}):
\begin{equation}
J = \frac{1}{4}\left(\sin 2x \sqrt{1+\sin^2 x} - \int \frac{2\sin^2 x-2\sin^4 x}
{\sqrt{1+\sin^2 x}}\, dx\right)+ \frac{1}{2}E(x|{-1}) \label{eqn:J2}
\end{equation}
\section{Some \emph{kombinowanie}}
We cast (\ref{eqn:J}) into an alternative form as follows:
\begin{align}
J/2 &= \frac{1}{2} \int \cos^2 x \sqrt{1+\sin^2 x} \, dx \\
&= \frac{1}{4} \int \frac{2 \cos^2 x (1 +\sin^2 x)}{\sqrt{1+\sin^2 x}} \, dx \\
&= \frac{1}{4} \int \frac{2(1-\sin^2 x)(1+\sin^2 x)}{\sqrt{1+\sin^2 x}} \, dx \\
&= \frac{1}{4} \int \frac{2-2\sin^4 x}{\sqrt{1+\sin^2 x}}\, dx \label{eqn:J3}
\end{align}
Adding (\ref{eqn:J3}) and (\ref{eqn:J2}) gives
\begin{align}
3J/2 &=
\frac{1}{4}\left(\sin 2x \sqrt{1+\sin^2 x} - \int \frac{2\sin^2 x-2\sin^4 x}
{\sqrt{1+\sin^2 x}}\, dx\right)+ \frac{1}{2}E(x|{-1})
+ \frac{1}{4}\int \frac{2-2\sin^4 x}{\sqrt{1+\sin^2 x}}\, dx \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + \frac{1}{2}E(x|{-1})
+ \frac{1}{4}\left(\int\frac{2-2\sin^4 x}{\sqrt{1+\sin^2 x}}\, dx
-\int\frac{2\sin^2 x-2\sin^4 x}{\sqrt{1+\sin^2 x}}\, dx\right) \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + \frac{1}{2}E(x|{-1})
+ \frac{1}{4}\left(\int\frac{2-2\sin^2 x}{\sqrt{1+\sin^2 x}}\, dx\right) \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + \frac{1}{2}E(x|{-1})
+ \frac{1}{4}\left(\int\frac{4}{\sqrt{1+\sin^2 x}}\, dx -
\int\frac{2+2\sin^2 x}{\sqrt{1+\sin^2 x}}\, dx\right) \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + \frac{1}{2}E(x|{-1})
+ \frac{1}{4}\left(\int\frac{4}{\sqrt{1+\sin^2 x}}\, dx -
\int 2\sqrt{1+\sin^2 x}\, dx\right) \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + \frac{1}{2}E(x|{-1})
+ \frac{1}{4}(4F(x|{-1}) - 2E(x|{-1})) \\
&= \frac{1}{4}\sin 2x \sqrt{1+\sin^2 x} + F(x|{-1}) \\
J &= \frac{1}{6}\sin 2x \sqrt{1+\sin^2 x} + \frac{2}{3}F(x|{-1})
\label{eqn:Jfinal}
\end{align}
\section{Home stretch}
Recall that $u = ix$, or $x = -iu$. Also, we see
that $\sin(ix) = (-i)i\sin(ix) = (-i)\sinh(i(ix)) = (-i)\sinh(-x) = i\sinh x$
by using (\ref{eqn:sinh}).
Now substitute (\ref{eqn:Jfinal}) back into (\ref{eqn:utox}), to obtain:
\begin{align}
\int \cosh^2 u \sqrt{1-\sinh^2 u} \, du
&= \frac{i}{6}\sin 2x \sqrt{1+\sin^2 x} + \frac{2i}{3}F(x|{-1}) \\
&= \frac{i}{6}\sin -2iu \sqrt{1+\sin^2 -iu} + \frac{2i}{3}F(-iu|{-1}) \\
&= \frac{i}{6}i\sinh -2u \sqrt{1+(i\sinh -u)^2} + \frac{2i}{3}F(-iu|{-1}) \\
&= \frac{1}{6}\sinh 2u \sqrt{1-\sinh^2 u} + \frac{2i}{3}F(-iu|{-1})
\label{eqn:coshfinal}
\end{align}
Finally, recall that $t=(-1)^{1/4}\sinh u$, so that
$u = \sinh^{-1}((-1)^{7/4}t)$. Substituting (\ref{eqn:coshfinal}) back into
(\ref{eqn:orig}) gives:
\begin{align}
\int \sqrt{1+t^4} \, dt &= (-1)^{1/4} \left(\frac{1}{6}\sinh 2u
\sqrt{1-\sinh^2 u} + \frac{2i}{3}F(-iu|{-1})\right) \\ \nonumber
&= (-1)^{1/4} \bigg(\frac{1}{6}\sinh[2\sinh^{-1}((-1)^{7/4}t)]
\sqrt{1-\sinh^2 (\sinh^{-1}((-1)^{7/4}t))} \\
&+ \frac{2i}{3}F(-i[\sinh^{-1}((-1)^{7/4}t)]|{-1})\bigg) \\ \nonumber
&= (-1)^{1/4} \bigg(\frac{1}{6}\cdot 2\sinh[\sinh^{-1}((-1)^{7/4}t)]
\cosh[\sinh^{-1}((-1)^{7/4}t)] \\ 
&\times \sqrt{1-((-1)^{7/4}t)^2} +
\frac{2i}{3}F(-i[\sinh^{-1}((-1)^{7/4}t)]|{-1})\bigg) \\ \nonumber
&= (-1)^{1/4} \bigg(\frac{1}{3}((-1)^{7/4}t)\sqrt{1+((-1)^{7/4}t)^2}
\sqrt{1-((-1)^{7/4})^2} \\ 
&+\frac{2i}{3}F(-i[\sinh^{-1}((-1)^{7/4}t)]|{-1})\bigg) \\
&= (-1)^{1/4} \left(\frac{1}{3}((-1)^{7/4}t)\sqrt{1-((-1)^{7/4}t)^4}
+\frac{2i}{3}F(-i[\sinh^{-1}((-1)^{7/4}t)]|{-1})\right) \\
&= \frac{1}{3}t\sqrt{1+t^4}+
\frac{2}{3}(-1)^{3/4}F(-i[\sinh^{-1}((-1)^{7/4}t)]|{-1}) \\
&= \frac{1}{3}t\sqrt{1+t^4}-
\frac{2}{3}(-1)^{1/4}F(i\sinh^{-1}((-1)^{1/4}t)|{-1})
\end{align}
which, when put over the denominator $3\sqrt{1+t^4}$, exactly matches
Mathematica's output.
\end{document}