April 18, 2025
We present a branch-consistent framework for integrals involving quadratic radicals by expressing exponentials of principal inverse trigonometric functions in simple algebraic forms. Two core identities for \(e^{\pm i\cos^{-1}(\cdot)}\) and \(e^{\pm i\sec^{-1}(\cdot)}\) (on principal branches) lead to five explicit substitutions (“Transforms”) that reduce common radical and half-angle integrands to rational functions of a single parameter. The method uniformly handles circular and hyperbolic cases, with differentials that do not depend on the \(\pm\) choice once branches are fixed, easing sign and domain bookkeeping. We recover Euler’s first and second substitutions from the transforms up to trivial reparametrizations and give worked examples; in particular, the classical Weierstrass substitution is obtained as a direct corollary of Transform 5 (unit-radius specialization), providing a clean illustration of the framework’s unifying reach. A short binomial-difference formula streamlines back-substitution expressions such as \(t^n-1/t^n\). The focus is methodological rather than exhaustive: the transforms are intended as compact templates that consolidate standard substitutions within a single, principal-branch calculus.
Classical techniques for integrating expressions with quadratic radicals—such as \(\sqrt{x^2-a^2}\) and \(\sqrt{a^2-x^2}\)—typically alternate between circular and hyperbolic substitutions. Euler’s first and second substitutions are standard examples: they rationalize integrands but require careful sign/branch tracking. The aim of this note is to give a compact, branch-consistent framework that unifies these substitutions and related trigonometric half–angle tricks under a single exponential parametrization. Everything is carried out on principal complex branches inherited from the principal logarithm and square root, so that the analytic behavior at endpoints and across domains is transparent and under precise control.
Our starting point is a pair of Euler-like identities 1 that express the exponentials of principal inverse trigonometric functions in simple algebraic forms. The first core result (Theorem 1) shows that, for \(y\in\mathbb{R}\), \[e^{-i\cos^{-1} y}=y-\sqrt{y^2-1}\quad(|y|\le1),\] \[e^{\pm i\cos^{-1} y}=\tan\!\Bigl(\tfrac12\,\csc^{-1}y\Bigr) =\frac{1-\tan\!\bigl(\tfrac12\,\sec^{-1}y\bigr)}{1+\tan\!\bigl(\tfrac12\,\sec^{-1}y\bigr)}\quad(|y|\ge1),\] with the signs fixed by the principal branches. The second core result (Theorem 2) provides the complementary identities with \(\sec^{-1}\) and \(\sin^{-1}\) interchanged, valid on both circular and hyperbolic regimes. A short “bridge” lemma (Lemma 1) connects the circular and hyperbolic parametrizations via \(z\mapsto i y\), yielding the familiar \(y+\sqrt{y^2+1}\) substitution directly from the same scheme.
From these identities we derive five explicit Transforms that turn integrals built from
\[\begin{align} &x,\quad \sqrt{(x+b)^2\pm a^2},\quad \sqrt{\frac{x+b-a}{x+b+a}},\quad \sqrt{\frac{a+b+x}{a-b-x}},\quad e^{\cos^{-1}(\cdot)},\quad e^{\sin^{-1}(\cdot)}, \\ &\tan\!\Bigl(\tfrac12\,\csc^{-1}(\cdot)\Bigr),\quad \tan\!\Bigl(\tfrac12\,\sec^{-1}(\cdot)\Bigr),\quad \tan\!\Bigl(\tfrac12\,\cos^{-1}(\cdot)\Bigr),\quad \tan\!\Bigl(\tfrac12\,\sin^{-1}(\cdot)\Bigr) \end{align}\]
into rational forms in a single parameter \(t=e^{\pm i\alpha}\) (circular), \(r=e^{\pm i\phi}\) (circular on \(|y|\le1\)), or \(s=e^{\theta}=y+\sqrt{y^2+1}\) (hyperbolic), with domain-consistent back–substitution formulas. The transforms come with built–in differentials \(dx\) that are independent of the \(\pm\) choice, so the same template handles both \(y\ge1\) and \(y\le-1\) (or \(|y|\le1\)) without additional casework. A compact binomial–difference identity (§5) further streamlines expressions like \(t^n-1/t^n\) that frequently arise after substitution.
A benchmark of 100 integrals involving mixed half-angle tangent composites demonstrates the practical efficiency of this unified approach. Using the piecewise back-substitution formula derived herein, Transform 1 outperforms Mathematica’s
Integrate in 82/100 cases, offering order-of-magnitude speedups on structurally complex integrands and significantly higher runtime predictability. The method mitigates expression swell: in our test set, the USM produced
“monster” antiderivatives (byte count \(\ge\) 10,000) in only 5 instances, compared to 24 for the general-purpose solver. [3]
One application underscores the scope and convenience of this unified substitution method. We show that the classical Weierstrass 2 substitution (for integrals rational in \(\sin\omega\) and \(\cos\omega\)) is an immediate corollary of Transform 5: in the unit-radius specialization, the USM parameter coincides with the half–angle parameter \(r=\tan(\omega/2)\). We also recover Euler’s first and second substitutions [8] from Transforms 2 and 5, with exact parameter correspondence up to the trivial reparametrizations \(t\leftrightarrow 1/t\) (Euler 1) and \(r=-t_E\) (Euler 2). The examples in §4 illustrate how routine radicals and half–angle composites collapse to rational integrals and how the principal-branch choices enforce the correct real limits.
In short, by organizing everything around \(e^{\pm i\cos^{-1}(\cdot)}\), \(e^{\pm i\sec^{-1}(\cdot)}\), and \(e^{\sinh^{-1}(\cdot)}\) on principal branches, we obtain a single, self-consistent calculus for circular and hyperbolic substitutions. This yields compact derivations, uniform domain handling, and subsumption of Euler’s classical substitutions by a common template.
Throughout, \(y\in\mathbb{R}\). We use the principal complex branches of the inverse trigonometric functions, inherited from the principal logarithm and square root: \[\log z \text{ is principal with cut }(-\infty,0],\quad \sqrt{\cdot}\text{ is the principal root (cut }(-\infty,0]) \cite{Ahlfors1979}.\] Define \[\sin^{-1} z := -\,i\log\!\bigl(iz+\sqrt{1-z^2}\bigr),\qquad \cos^{-1} z := \frac{\pi}{2}-\sin^{-1} z,\] and \[\sec^{-1} z := \cos^{-1}\!\bigl(1/z\bigr),\qquad \csc^{-1} z := \sin^{-1}\!\bigl(1/z\bigr),\] with the usual understanding that \(\sec^{-1} 0\) and \(\csc^{-1} 0\) are undefined, and limits at the boundary points are taken when explicitly indicated [9]. For real \(y\) in the classical domains these agree with the standard real-valued choices: \[\cos^{-1} y\in[0,\pi],\quad \sin^{-1} y\in\bigl[-\tfrac{\pi}{2},\tfrac{\pi}{2}\bigr],\] and \[\sec^{-1} y\in[0,\pi]\setminus\{\tfrac{\pi}{2}\},\quad \csc^{-1} y\in\bigl(-\tfrac{\pi}{2},0\bigr)\cup\bigl(0,\tfrac{\pi}{2}\bigr]. \cite{AS1964}\] In particular, for \(|y|<1\) the principal square roots satisfy \[\sqrt{y^2-1}=i\,\sqrt{1-y^2}\qquad(\Im\sqrt{y^2-1}\ge0). \cite{Ahlfors1979}\]
Theorem 1. Let \[\alpha=\cos^{-1}(y),\qquad \beta=\csc^{-1}(y),\qquad \gamma=\sec^{-1}(y).\] Then, with principal square roots throughout:
If \(|y|\le 1\), then \[e^{-i\alpha}=y-\sqrt{y^2-1}\qquad\text{with }\;\sqrt{y^2-1}=i\sqrt{1-y^2}.\]
If \(|y|\ge 1\), then \[e^{\pm i\alpha}=\tan\!\Bigl(\frac{\beta}{2}\Bigr) =\frac{1-\tan\!\bigl(\frac{\gamma}{2}\bigr)}{\,1+\tan\!\bigl(\frac{\gamma}{2}\bigr)}\,,\] with the upper sign when \(y\ge 1\) and the lower sign when \(y\le -1\). (Remark: the expression in \(\gamma\) is undefined at \(y=-1\) since \(\tan(\gamma/2)\) blows up; equality at \(y=-1\) holds by limit.)
Proof. (A) For \(|y|\le1\), \(\alpha=\cos^{-1} y\in[0,\pi]\) gives \(\cos\alpha=y\) and \(\sin\alpha=\sqrt{1-y^2}\ge0\). Hence \[e^{-i\alpha}=\cos\alpha-i\sin\alpha = y-i\sqrt{1-y^2} = y-\sqrt{y^2-1},\] using \(\sqrt{y^2-1}=i\sqrt{1-y^2}\) for the principal root.
(B) Assume \(|y|\ge1\). First, with \(\beta=\csc^{-1} y\) we have \(\sin\beta=1/y\) and \[\tan\!\Bigl(\frac{\beta}{2}\Bigr)=\frac{1-\cos\beta}{\sin\beta} =\frac{1-\sqrt{1-1/y^2}}{1/y} = \begin{cases} y-\sqrt{y^2-1}, & y\ge 1,\\[4pt] y+\sqrt{y^2-1}, & y\le -1, \end{cases}\] where \(\sqrt{1-1/y^2}=\dfrac{\sqrt{y^2-1}}{|y|}\) on principal branches.
Meanwhile \(\alpha=\cos^{-1} y\) gives \[e^{\pm i\alpha}=\cos\alpha\pm i\sin\alpha = y\pm i\sqrt{1-y^2} = y\mp\sqrt{y^2-1},\] since \(\sqrt{1-y^2}=i\sqrt{y^2-1}\). Thus \(e^{+i\alpha}=y-\sqrt{y^2-1}\) for \(y\ge1\) and \(e^{-i\alpha}=y+\sqrt{y^2-1}\) for \(y\le-1\), matching the sign split above.
For the alternative expression via \(\gamma=\sec^{-1} y\) (so \(\cos\gamma=1/y\), \(\sin\gamma=\sqrt{1-1/y^2}=\dfrac{\sqrt{y^2-1}}{|y|}\)), use the half-angle identity \[\tan\!\Bigl(\frac{\gamma}{2}\Bigr)=\frac{\sin\gamma}{1+\cos\gamma}. \cite{Hobson1928}\] When \(y\ge1\), \[\tan\!\Bigl(\frac{\gamma}{2}\Bigr) =\frac{\frac{\sqrt{y^2-1}}{y}}{1+\frac{1}{y}} =\frac{\sqrt{y^2-1}}{y+1},\] and then \[\frac{1-\tan(\gamma/2)}{1+\tan(\gamma/2)} =\frac{1-\frac{\sqrt{y^2-1}}{y+1}}{1+\frac{\sqrt{y^2-1}}{y+1}} =\frac{y+1-\sqrt{y^2-1}}{y+1+\sqrt{y^2-1}} = y-\sqrt{y^2-1}.\] When \(y\le-1\), \[\tan\!\Bigl(\frac{\gamma}{2}\Bigr) =\frac{\frac{\sqrt{y^2-1}}{|y|}}{1+\frac{1}{y}} =-\,\frac{\sqrt{y^2-1}}{\,y+1\,},\] and hence \[\frac{1-\tan(\gamma/2)}{1+\tan(\gamma/2)} =\frac{1+\frac{\sqrt{y^2-1}}{y+1}}{1-\frac{\sqrt{y^2-1}}{y+1}} =\frac{y+1+\sqrt{y^2-1}}{y+1-\sqrt{y^2-1}} = y+\sqrt{y^2-1}.\] Combining with the earlier identification of \(e^{\pm i\alpha}\) completes the proof. (At \(y=-1\), interpret the \(\gamma\)-formula by continuity.) ◻
Theorem 2. Let \[\phi=\sec^{-1}(y),\qquad \psi=\sin^{-1}(y),\qquad \alpha=\cos^{-1}(y).\] Then, with principal square roots throughout:
If \(|y|\le 1\) and \(y\neq 0\), then \[e^{\pm i\phi}=\tan\!\Bigl(\frac{\psi}{2}\Bigr),\] with the \(+\) sign for \(y\in[0,1]\) and the \(-\) sign for \(y\in[-1,0]\). (At \(y=0\), \(\sec^{-1} 0\) is undefined; the identity holds by the limit \(y\to0^\pm\).)
If \(|y|\ge 1\), then \[e^{\pm i\phi}=\tan\!\Bigl(\frac{\psi}{2}\Bigr),\] with the \(-\) sign when \(y\ge 1\) and the \(+\) sign when \(y\le -1\).
(Equivalent form on \(|y|\le 1\) with \(y\neq0\)) \[e^{\pm i\phi}=\frac{1-\tan\!\bigl(\frac{\alpha}{2}\bigr)}{\,1+\tan\!\bigl(\frac{\alpha}{2}\bigr)}\,,\] with the \(+\) sign for \(y\in[0,1]\) and the \(-\) sign for \(y\in[-1,0]\). (Again, interpret \(y=0\) by limit.)
Proof. (A) First take \(y\in[0,1]\). Write \(y=\operatorname{sech}u\) with \(u\ge0\). Then \(\phi=iu\) satisfies \(\sec\phi=\operatorname{sech}u\), so \(e^{i\phi}=e^{-u}\). With \(\psi=\sin^{-1} y\) we have \[\sin\psi=\operatorname{sech}u,\qquad \cos\psi=\tanh u,\] and hence \[\tan\!\Bigl(\frac{\psi}{2}\Bigr)=\frac{1-\cos\psi}{\sin\psi} =\frac{1-\tanh u}{\operatorname{sech}u} =\cosh u-\sinh u =e^{-u}=e^{\,i\phi}.\]
For \(y\in[-1,0]\), write \(y=-\operatorname{sech}u\) with \(u\ge0\). Then \(\phi=\pi-iu\) gives \(\sec\phi=y\), so \[e^{-i\phi}=e^{-i(\pi-iu)}=e^{-i\pi}\,e^{-u}=-\,e^{-u}.\] Also \(\psi=\sin^{-1}(y)=-\sin^{-1}(\operatorname{sech}u)\), hence \[\tan\!\Bigl(\frac{\psi}{2}\Bigr) =-\,\frac{1-\tanh u}{\operatorname{sech}u} =-(\cosh u-\sinh u) =-e^{-u}=e^{-i\phi},\] which is the claimed sign choice.
(B) For \(|y|\ge1\), take principal square roots so that \(\sqrt{1-y^2}=i\sqrt{y^2-1}\) with \(\Im\ge0\). Since \(\psi=\sin^{-1} y\) has \(\sin\psi=y\) and \(\cos\psi=\sqrt{1-y^2}\), \[\tan\!\Bigl(\frac{\psi}{2}\Bigr) =\frac{1-\cos\psi}{\sin\psi} =\frac{1-i\sqrt{y^2-1}}{y}.\] With \(\phi=\sec^{-1} y\), we have \[\cos\phi=\frac{1}{y},\qquad \sin\phi=\sqrt{1-\frac{1}{y^2}}=\frac{\sqrt{y^2-1}}{|y|}.\] Thus \[e^{-i\phi}=\cos\phi-i\sin\phi=\frac{1}{y}-i\,\frac{\sqrt{y^2-1}}{|y|} =\begin{cases} \dfrac{1-i\sqrt{y^2-1}}{y}, & y\ge1,\\[6pt] \dfrac{1+i\sqrt{y^2-1}}{y}, & y\le-1, \end{cases}\] and likewise \[e^{+i\phi}=\cos\phi+i\sin\phi =\begin{cases} \dfrac{1+i\sqrt{y^2-1}}{y}, & y\ge1,\\[6pt] \dfrac{1-i\sqrt{y^2-1}}{y}, & y\le-1. \end{cases}\] Comparing with \(\tan(\psi/2)=\dfrac{1-i\sqrt{y^2-1}}{y}\) gives the stated sign rule: \(e^{-i\phi}\) for \(y\ge1\) and \(e^{+i\phi}\) for \(y\le-1\).
(C) On \(|y|\le1\), let \(\alpha=\cos^{-1} y\) so that \(\cos\alpha=y\) and \(\sin\alpha=\sqrt{1-y^2}\). Using the half-angle identity [10] \[\tan\!\Bigl(\frac{\alpha}{2}\Bigr)=\frac{1-\cos\alpha}{\sin\alpha}=\frac{1-y}{\sqrt{1-y^2}},\] we compute \[\frac{1-\tan(\alpha/2)}{1+\tan(\alpha/2)} =\frac{1-\frac{1-y}{\sqrt{1-y^2}}}{1+\frac{1-y}{\sqrt{1-y^2}}} =\frac{\sqrt{1-y^2}-(1-y)}{\sqrt{1-y^2}+(1-y)}.\] For \(y\in[0,1]\), the hyperbolic substitution \(y=\operatorname{sech}u\) gives \[\frac{\sqrt{1-y^2}-(1-y)}{\sqrt{1-y^2}+(1-y)} = \frac{\tanh u-(1-\operatorname{sech}u)}{\tanh u+(1-\operatorname{sech}u)} = e^{-u}=e^{\,i\phi}.\] For \(y\in[-1,0]\), with \(y=-\operatorname{sech}u\) we similarly obtain \(e^{-i\phi}\). This matches the sign split claimed. ◻
Throughout let \(a>0\), \(b\in\mathbb{R}\), and \[y:=\frac{x+b}{a}.\] Principal branches are as in the “Conventions” section. We repeatedly use \[\cos\alpha=\frac{t+t^{-1}}{2},\quad \sin\alpha= \begin{cases} \dfrac{t-t^{-1}}{2i}, & t=e^{+i\alpha},\\[6pt] \dfrac{t^{-1}-t}{2i}, & t=e^{-i\alpha}, \end{cases} \quad\text{when }t=e^{\pm i\alpha},\] \[\cosh\theta=\frac{s+s^{-1}}{2},\;\sinh\theta=\frac{s-s^{-1}}{2}\;\text{when }s=e^\theta.\]
Lemma 1 (Bridge lemma). For \(y\in\mathbb{R}\), on principal branches, \[e^{\,i\cos^{-1}(iy)} \;=\; i\!\left(y+\sqrt{y^{2}+1}\right), \qquad\text{equivalently}\qquad -\,i\,e^{\,i\cos^{-1}(iy)} \;=\; y+\sqrt{y^{2}+1}.\]
Proof. We work on principal branches: \(\log\) and \(\sqrt{\phantom{x}}\) use the principal cut \((-\infty,0]\), and the inverse trigonometric functions inherit branches from the principal \(\log\). For real \(y\), \[\sin^{-1}(iy)=i\,\sinh^{-1}(y), \qquad \sinh^{-1}(y)=\log\!\bigl(y+\sqrt{y^2+1}\bigr). \cite{AS1964}\] Since \(\cos^{-1}z=\tfrac{\pi}{2}-\sin^{-1}z\) on these branches, with \(z=iy\) we obtain \[\begin{align} e^{\,i\cos^{-1}(iy)} &= e^{\,i\!\left(\frac{\pi}{2}-\sin^{-1}(iy)\right)}\\ &= i\,e^{-\,i\,\sin^{-1}(iy)}\\ &= i\,e^{-\,i\cdot i\,\sinh^{-1}(y)}\\ &= i\,e^{\,\sinh^{-1}(y)}\\ &= i\!\left(y+\sqrt{y^{2}+1}\right). \end{align}\] as claimed. Multiplying both sides by \(-i\) yields the equivalent form. ◻
Let \[\alpha=\cos^{-1}(y),\qquad \beta=\csc^{-1}(y),\qquad \gamma=\sec^{-1}(y).\] Set \[t:=e^{\pm i\alpha} \quad\text{with the sign chosen by Theorem~\ref{thm:cos-core}(B)}.\] Then (Theorem 1) \[\tan\!\Bigl(\frac{\beta}{2}\Bigr)=t, \qquad \tan\!\Bigl(\frac{\gamma}{2}\Bigr)=\frac{1-t}{1+t}.\] Moreover, \[x=a\cos\alpha-b=a\frac{t+t^{-1}}{2}-b=\frac{a(t^2+1)}{2t}-b,\qquad dx=-a\sin\alpha\,d\alpha = a\,\frac{t^2-1}{2t^2}\,dt,\] where the final formula for \(dx\) is independent of the \(\pm\) choice (it cancels). Hence \[\boxed{\; \int f\!\left(x,\,\tan\frac{\beta}{2},\,\tan\frac{\gamma}{2}\right)\,dx = \int f\!\left(\frac{a(t^2+1)}{2t}-b,\,t,\frac{1-t}{1+t}\right)\, a\,\frac{t^2-1}{2t^2}\,dt. \;}\] Use when \(\lvert y\rvert\ge 1\). Fix the branch by \[t= \begin{cases} y-\sqrt{y^2-1}, & y\ge 1,\\[4pt] y+\sqrt{y^2-1}, & y\le -1, \end{cases}\] and back-substitute \[t= \begin{cases} \dfrac{x+b-\sqrt{(x+b)^2-a^2}}{a}, & y\ge 1,\\[10pt] \dfrac{x+b+\sqrt{(x+b)^2-a^2}}{a}, & y\le -1. \end{cases}\]
With the same \(t=e^{\pm i\alpha}\) as above, \[\sqrt{(x+b)^2-a^2} =a\,\sqrt{\cos^2\alpha-1} =\pm\,i\,a\sin\alpha =\pm\,a\,\frac{t-\tfrac{1}{t}}{2} =\pm\,a\,\frac{t^2-1}{2t},\] and (using the half-angle relation from Theorem 1(B)) \[\sqrt{\frac{x+b-a}{x+b+a}} =\sqrt{\frac{y-1}{y+1}} =\frac{1-t}{1+t}.\] Together with \(x=\dfrac{a(t^2+1)}{2t}-b\) and \(dx=a\,\dfrac{t^2-1}{2t^2}\,dt\), this yields \[\boxed{\, \int f\!\left(\begin{array}{@{}l@{}} x,\\[2pt] \sqrt{(x+b)^2-a^2},\\[2pt] \sqrt{\dfrac{x+b-a}{x+b+a}} \end{array}\right)\,dx \;=\; \int f\!\left(\begin{array}{@{}l@{}} \dfrac{a(t^2+1)}{2t}-b,\\[2pt] \mp\,a\,\dfrac{t^2-1}{2t},\\[2pt] \dfrac{1-t}{1+t} \end{array}\right)\;\; a\,\dfrac{t^2-1}{2t^2}\,dt \,}\] Use when \(\lvert y\rvert\ge 1\). Take \(t\) exactly as in Transform 1. Here \[\sqrt{(x+b)^2-a^2}= \begin{cases} -\,a\,\dfrac{t^2-1}{2t}, & y\ge 1,\\[8pt] \;\;a\,\dfrac{t^2-1}{2t}, & y\le -1. \end{cases}\] Back-substitution for \(t\) is the same as in Transform 1.
Let \(\theta=\sinh^{-1}(y)\), \(s:=e^\theta\) so that \[y=\sinh\theta=\frac{s-s^{-1}}{2},\qquad \sqrt{y^2+1}=\cosh\theta=\frac{s+s^{-1}}{2}.\] By Lemma 1, \(s=e^\theta=\sinh\theta+\cosh\theta=y+\sqrt{y^2+1}=-i\,e^{\,i\cos^{-1}(iy)}\) [9], so this hyperbolic substitution is itself generated from Theorem 1. Then \[x=a\sinh\theta-b=a\frac{s-s^{-1}}{2}-b=\frac{a(s^2-1)}{2s}-b,\] \[\sqrt{(x+b)^2+a^2}=a\cosh\theta=a\frac{s+s^{-1}}{2}=\frac{a(s^2+1)}{2s},\] and \(dx=a\cosh\theta\,d\theta=a\,\dfrac{s^2+1}{2s^2}\,ds\). Hence \[\boxed{\; \int f\!\left(x,\;\sqrt{(x+b)^2+a^2}\right)\,dx = \int f\!\left(\frac{a(s^2-1)}{2s}-b,\;\frac{a(s^2+1)}{2s}\right) a\,\frac{s^2+1}{2s^2}\,ds \;}.\] Back-substitute with \(s=e^\theta=y+\sqrt{y^2+1}=\dfrac{x+b+\sqrt{(x+b)^2+a^2}}{a}\).
Let \[\phi=\sec^{-1}(y),\qquad \psi=\sin^{-1}(y),\qquad \alpha=\cos^{-1}(y).\] By Theorem 2(A,C), set \[r:=e^{\pm i\phi}=\tan\!\Bigl(\frac{\psi}{2}\Bigr), \qquad r=\frac{1-\tan(\alpha/2)}{1+\tan(\alpha/2)}.\] Solving gives \[\tan\!\Bigl(\frac{\alpha}{2}\Bigr)=\frac{1-r}{1+r}\quad \text{(the \pm in r=e^{\pm i\phi} is carried in the choice of branch for r)}.\] Also, since \(y=\sec\phi\), one gets \[y=\frac{2r}{1+r^2},\qquad x=\frac{2ar}{1+r^2}-b,\qquad dx=a\cdot \frac{2(1-r^2)}{(1+r^2)^2}\,dr.\] Therefore \[\boxed{\; \int f\!\left(x,\,\tan\frac{\psi}{2},\,\tan\frac{\alpha}{2}\right)\,dx = \int f\!\left(\frac{2ar}{1+r^2}-b,\,r,\,\,\frac{1-r}{1+r}\right)\, a\,\frac{2(1-r^2)}{(1+r^2)^2}\,dr \;}.\] Use when \(\lvert y\rvert\le 1\). Fix \[\begin{align} r &= \tan\!\Bigl(\frac{\psi}{2}\Bigr) = \frac{y}{1+\sqrt{1-y^{2}}}\\ &= \frac{x+b}{\,a+\sqrt{a^{2}-(x+b)^{2}}\,}\\ &= \begin{cases} \dfrac{a-\sqrt{a^{2}-(x+b)^{2}}}{\,x+b\,}, & x\neq -b,\\ 0, & x=-b \text{ (by continuity).} \end{cases}\\ &\text{(\lvert x+b\rvert \le a, principal real root)} \end{align}\] Back-substitute with this \(r\).
Keep the same \(r=e^{\pm i\phi}=\tan(\psi/2)\) as above. Then \[\sqrt{a^2-(x+b)^2}=a\sqrt{1-y^2} =a\,\frac{1-r^2}{1+r^2}\quad(\text{principal branches}),\] and \[\sqrt{\frac{a+b+x}{a-b-x}} =\sqrt{\frac{1+y}{1-y}} =\frac{1+r}{1-r}.\] With \(x=\dfrac{2ar}{1+r^2}-b\) and \(dx=a\,\dfrac{2(1-r^2)}{(1+r^2)^2}\,dr\), we obtain \[\boxed{\,\int f\!\left(\begin{array}{@{}l@{}} x,\\[2pt] \sqrt{a^{2}-(x+b)^{2}},\\[2pt] \sqrt{\dfrac{a+b+x}{\,a-b-x\,}} \end{array}\right)\,dx \;=\; \int f\!\left(\begin{array}{@{}l@{}} \dfrac{2ar}{1+r^{2}}-b,\\[2pt] a\,\dfrac{1-r^{2}}{1+r^{2}},\\[2pt] \dfrac{1+r}{1-r} \end{array}\right)\; a\,\dfrac{2(1-r^{2})}{(1+r^{2})^{2}}\,dr \,}\] Use when \(\lvert y\rvert\le 1\). Take \(r\) exactly as in Transform 4 (and back-substitute with that same formula).
A classical target is an integral of the form \[I=\int R(\sin\omega,\cos\omega)\,d\omega,\] where \(R\) is rational in its arguments. To connect this with Transform 5, first reduce to a circular radical in the variable \(x\) by setting \(x=\sin\omega\). On any interval where the principal square root agrees with \(\cos\omega\) (for instance \(\omega\in[-\pi/2,\pi/2]\), so \(\cos\omega\ge0\)), we have \[\cos\omega=\sqrt{1-x^{2}},\qquad d\omega=\frac{dx}{\sqrt{1-x^{2}}},\] and hence \[I=\int \frac{R\!\left(x,\sqrt{1-x^{2}}\right)}{\sqrt{1-x^{2}}}\,dx.\]
Now apply Transform 5 with \(a=1\) and \(b=0\) (so \(y=x\) and \(|y|\le1\)). With \(r\) as in Transform 4–5, i.e.\(r=\tan(\psi/2)\) where \(\psi=\sin^{-1}(x)\), Transform 5 gives \[x=\frac{2r}{1+r^{2}},\qquad \sqrt{1-x^{2}}=\frac{1-r^{2}}{1+r^{2}},\qquad dx=\frac{2(1-r^{2})}{(1+r^{2})^{2}}\,dr.\] Substituting into the reduced integral yields the rational form \[\begin{align} I &=\int \frac{R\!\left(\frac{2r}{1+r^{2}},\frac{1-r^{2}}{1+r^{2}}\right)}{\frac{1-r^{2}}{1+r^{2}}}\cdot \frac{2(1-r^{2})}{(1+r^{2})^{2}}\,dr \\ &=\int R\!\left(\frac{2r}{1+r^{2}},\frac{1-r^{2}}{1+r^{2}}\right)\, \frac{2}{1+r^{2}}\,dr. \end{align}\] On the principal branch where \(\psi=\sin^{-1}(x)=\omega\), this parameter is \[r=\tan\!\Bigl(\frac{\psi}{2}\Bigr)=\tan\!\Bigl(\frac{\omega}{2}\Bigr),\] and the last display is exactly the classical Weierstrass substitution.
Relation to Euler 2. In §6 we showed that, in the circular case, Euler’s second parameter \(t_E\) satisfies \(t_E=-r\) (up to the harmless sign convention noted there). Thus, for the unit-radius specialization \(a=1\), \(b=0\), the Weierstrass half–angle parameter \(\tan(\omega/2)\) is precisely the USM parameter \(r\), i.e.the negative of Euler 2’s parameter under the same normalization.
Remark 1. In many cases, the USM maps the basic radicals to simple factors in the parameter together with a Jacobian that carries the same factors:
Transform 5: \[\sqrt{a^{2}-(x+b)^{2}}\;\mapsto\;a\frac{1-r^{2}}{1+r^{2}},\quad \sqrt{\frac{a+b+x}{\,a-b-x\,}}\;\mapsto\;\frac{1+r}{1-r},\quad dx\;\mapsto\;a\,\frac{2(1-r^{2})}{(1+r^{2})^{2}}\,dr.\]
Transform 2: \[\sqrt{(x+b)^{2}-a^{2}}\;\mapsto\;\pm\,a\,\frac{t^{2}-1}{2t},\quad \sqrt{\frac{x+b-a}{\,x+b+a\,}}\;\mapsto\;\frac{1-t}{1+t},\quad dx\;\mapsto\;a\,\frac{t^{2}-1}{2t^{2}}\,dt.\]
By quick inspection, if the factors \((1\pm r)\), \((1+r^{2})\) (resp.\((1\pm t)\)) that arise from the radicals are already present in the integrand so that the Jacobian’s copies cancel them, the transformed integrand collapses to a polynomial-type integrand in the parameter. Then the integral reduces to term-by-term antiderivatives (e.g., a combination of \(r^{-3}, r^{-1}, r\)), typically faster than ad-hoc trigonometric substitutions, partial fractions or Hermite reduction. For worked instances of this cancellation, see Example 6 (in \(r\), under Transform 5) and Example 4 (in \(t\), under Transform 2).
Throughout, \(a>0\), \(b\in\mathbb{R}\), and \(y=\dfrac{x+b}{a}\); principal branches are as in the Conventions. We refer to Theorems 1–2 and Transforms 1–5 from the previous section.
Here \(a=1\), \(b=0\), so \(y=x\) and \(|y|\ge1\). Apply Transform 2 with the upper-sign branch (since \(x\ge1\)), i.e. \[t=e^{+i\alpha}=y-\sqrt{y^2-1}=x-\sqrt{x^2-1},\qquad \sqrt{x^2-1}=-\,\frac{t^2-1}{2t},\quad dx=\frac{t^2-1}{2t^2}\,dt.\] Thus \[\int\sqrt{x^2-1}\,dx=\int\!\Bigl(-\frac{t^2-1}{2t}\Bigr)\,\frac{t^2-1}{2t^2}\,dt =-\int\frac{(t^2-1)^2}{4t^3}\,dt,\] which integrates to \[\frac{1}{2}\!\left(x\sqrt{x^2-1}-\log\!\bigl(x+\sqrt{x^2-1}\bigr)\right)+C.\] For \(x\le-1\) take the lower-sign branch in Transform 2.
Complete the square: \(\sqrt{x^2+x}=\sqrt{\,(x+\tfrac12)^2-\tfrac14\,}\), so take \(a=\tfrac12\), \(b=\tfrac12\) and use Transform 2 with the appropriate sign (upper sign if \(x>0\)). Then \[x=\frac{a(t^2+1)}{2t}-b=\frac{t^2+1}{4t}-\frac{1}{2},\quad \sqrt{x^2+x}=-\,\frac{a(t^2-1)}{2t}=-\,\frac{t^2-1}{4t},\quad dx=\frac{t^2-1}{4t^2}\,dt.\] Hence \[\int\frac{dx}{x\sqrt{x^2+x}} =-\,4\!\int\frac{dt}{(t-1)^2} =\frac{4}{t-1}+C.\] With \(t=2x+1-2\sqrt{x^2+x}\) for \(x>0\) (upper sign), this simplifies to \[\int \frac{dx}{x\sqrt{x^2+x}}=\frac{2}{x-\sqrt{x^2+x}}+C.\] Use the conjugate branch for \(x<-1\).
Use Transform 3 (hyperbolic): set \(\theta=\sinh^{-1}x\), \(s=e^\theta=x+\sqrt{1+x^2}\) so \(dx=\dfrac{s^2+1}{2s^2}\,ds\). Limits \(x:0\to\infty\) map to \(s:1\to\infty\). Then \[\int_{0}^{\infty}\frac{dx}{\bigl(x+\sqrt{1+x^2}\bigr)^2} =\int_{1}^{\infty}\frac{1}{s^2}\cdot\frac{s^2+1}{2s^2}\,ds =\frac{1}{2}\int_{1}^{\infty}\Bigl(s^{-2}+s^{-4}\Bigr)\,ds=\frac{2}{3}.\]
Note \(\dfrac{x+1}{x+3}=\dfrac{x+b-a}{x+b+a}\) with \(a=1\), \(b=2\). Apply Transform 2 (upper sign for \(x\ge-1\)): \[\sqrt{\frac{x+1}{x+3}}=\frac{1-t}{1+t},\quad dx=\frac{t^2-1}{2t^2}\,dt,\quad t=x+2-\sqrt{x^2+4x+3}.\] Thus \[\int \sqrt{\frac{x+1}{x+3}}\,dx =\int\frac{1-t}{1+t}\cdot\frac{t^2-1}{2t^2}\,dt =-\frac{1}{2}\int\!\Bigl(1-2t^{-1}+t^{-2}\Bigr)\,dt =\ln|t|+\frac{1}{2}\Bigl(t^{-1}-t\Bigr)+C,\] hence \[\int \sqrt{\frac{x+1}{x+3}}\,dx =\ln\,\!\bigl(x+2-\sqrt{x^2+4x+3}\bigr)+\sqrt{x^2+4x+3}+C.\]
Remark 2. It is instructive to contrast the algebraic economy of the USM with standard approaches for this integrand. The classical rationalization \[u=\sqrt{\frac{x+1}{x+3}}\] yields the rational form \(\int \frac{4u^2}{(u^2-1)^2}\,du\), which typically necessitates a cumbersome partial fraction decomposition. Attempting to bypass this via a secondary substitution introduces its own friction: while the hyperbolic choice \(u=\coth z\) leads to a manageable integration of \(\cosh^2 z\), the subsequent back-substitution is algebraically tedious, requiring double-angle expansions and inverse identities to revert to \(x\). Similarly, the trigonometric choice \(u=\sec \theta\) results in the laborious \(\int \csc^3 \theta\,d\theta\) (requiring recursive integration by parts or reduction formulas that few people remember by heart). Crucially, both traditional paths impose a distinct second layer of substitution (\(x \to u \to z\) or \(\theta\)), whereas USM Transform 2 structurally cancels the denominator in a single step, collapsing the integrand immediately to the elementary \(1 - 2t^{-1} + t^{-2}\).
On \(|x|\ge1\) use Transform 1 4: with \(\alpha=\cos^{-1}(y)\), \(\beta=\csc^{-1}(y)\), \(\gamma=\sec^{-1}(y)\), \[t=e^{\pm i\alpha}=\tan\!\Bigl(\frac{\beta}{2}\Bigr),\qquad \frac{1-\tan(\gamma/2)}{1+\tan(\gamma/2)}=t.\] For \(x\ge1\) take the upper sign; then \[\frac{1}{\tan(\beta/2)-\tan(\gamma/2)}\;dx =\frac{1+t}{t^2+2t-1}\cdot\frac{t^2-1}{2t^2}\,a\,dt = \frac{a}{2}\,\frac{(t-1)(t+1)^2}{t^2(t^2+2t-1)}\,dt,\] a rational integrand in \(t\) amenable to partial fractions. For \(x\le-1\) use the lower-sign branch.
Set \(a=2\), \(b=0\) and apply Transform 5 on \(|y|\le1\) with \[r=e^{+i\phi}=\tan\!\Bigl(\frac{\psi}{2}\Bigr)=\frac{2-\sqrt{4-x^2}}{x},\quad x=\frac{4r}{1+r^2},\quad \sqrt{4-x^2}=\frac{2(1-r^2)}{1+r^2},\] and \[dx=\frac{4(1-r^2)}{(1+r^2)^2}\,dr.\] Substituting gives \[\int\frac{dx}{x^3\sqrt{4-x^2}} =\frac{1}{32}\int\!\Bigl(r^{-3}+2r^{-1}+r\Bigr)\,dr =\frac{1}{16}\ln|r|+\frac{1}{64}\!\left(r^2-\frac{1}{r^2}\right)+C,\] i.e. \[\int\frac{dx}{x^3\sqrt{4-x^2}} =\frac{1}{16}\!\left(\ln\left|\!\frac{2-\sqrt{4-x^2}}{x}\right|-\frac{2\sqrt{4-x^2}}{x^2}\right)+C.\]
Recall the half–angle identity \[\tan\!\Bigl(\tfrac12\,\csc^{-1}(x^{2})\Bigr)=x^{2}-\sqrt{x^{4}-1}\qquad(|x|\ge1).\]
With the substitution \(u=x^{2}\) and the restriction \(x\ge1\) (so \(du=2x\,dx\) and \(dx=\dfrac{du}{2\sqrt{u}}\)), the integral becomes \[\frac{1}{2}\int \frac{\sqrt{\,u-\sqrt{u^{2}-1}\,}}{\sqrt{u}}\;du.\]
Following the USM (Transform 2, upper sign), set \[t:=u-\sqrt{u^{2}-1}\qquad (t>0,\;u\ge 1).\]
Then \[u=\frac{t+t^{-1}}{2},\qquad \sqrt{u^{2}-1}=\frac{t^{-1}-t}{2},\qquad du=\frac{t^{2}-1}{2t^{2}}\,dt,\] and \[\sqrt{u}=\frac{\sqrt{t^{2}+1}}{\sqrt{2}\,t^{1/2}}, \qquad \sqrt{\,u-\sqrt{u^{2}-1}\,}=\sqrt{t}.\]
Therefore \[\frac{1}{2}\int \frac{\sqrt{\,u-\sqrt{u^{2}-1}\,}}{\sqrt{u}}\;du =\frac{1}{2}\int \Bigl(\sqrt{2}\,\frac{t}{\sqrt{t^{2}+1}}\Bigr)\Bigl(\frac{t^{2}-1}{2t^{2}}\Bigr)\,dt =\frac{\sqrt{2}}{4}\!\int\!\Bigl(\frac{t}{\sqrt{t^{2}+1}}-\frac{1}{t\sqrt{t^{2}+1}}\Bigr)dt.\]
Integrating term-by-term gives \[\int \sqrt{\tan\!\Bigl(\tfrac12\,\csc^{-1}(x^{2})\Bigr)}\,dx =\frac{\sqrt{2}}{4}\!\left( \sqrt{t^{2}+1} -\ln\!\left|\frac{t+\sqrt{t^{2}+1}-1}{t+\sqrt{t^{2}+1}+1}\right| \right)+C.\]
Finally, since \(t=\tan\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)\) and \(\sqrt{t^{2}+1}=\sec\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)\), the antiderivative can be written as \[\frac{\sqrt{2}}{4}\,\sec\!\Bigl(\tfrac12\,\csc^{-1}(x^{2})\Bigr) -\frac{\sqrt{2}}{4}\,\ln\!\left| \frac{\tan\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)+\sec\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)-1}{\tan\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)+\sec\!\bigl(\tfrac12\,\csc^{-1}(x^{2})\bigr)+1} \right|+C.\]
Using \[\frac{\tan\omega+\sec\omega-1}{\tan\omega+\sec\omega+1}=\tan\!\Bigl(\frac{\omega}{2}\Bigr),\] with \(\omega=\tfrac12\,\csc^{-1}(x^{2})\), we obtain the compact form \[\int \sqrt{\tan\!\Bigl(\tfrac12\,\csc^{-1}(x^{2})\Bigr)}\,dx =\frac{\sqrt{2}}{4}\left[ \sec\!\Bigl(\tfrac12\,\csc^{-1}(x^{2})\Bigr) -\ln\Bigl|\tan\!\Bigl(\tfrac14\,\csc^{-1}(x^{2})\Bigr)\Bigr| \right]+C.\]
Let \(\alpha=\cos^{-1}x\in[0,\pi]\) and set \[t:=e^{-i\alpha}=x-i\sqrt{1-x^2}\qquad\text{(Theorem~\ref{thm:cos-core}(A), principal branches).}\] Then, on the principal branch, \(\log t=-i\alpha\) holds for \(\alpha\in[0,\pi)\) (equivalently \(x\in(-1,1]\)), since \(t=e^{-i\alpha}\) stays off the branch cut of \(\log\). At the endpoint \(x=-1\) (so \(\alpha=\pi\) and \(t=-1\) lies on the branch cut), the identity is interpreted by continuity. Moreover, because the integrand involves \(t^{\,i}=e^{\,i\log t}\), changing the branch of \(\log\) would multiply \(t^{\,i}\) by a nontrivial factor \(e^{-2\pi k}\) rather than merely altering the antiderivative by an additive constant. Hence
\[e^{\cos^{-1}x}=e^{\alpha}=e^{\,i\log t}=t^{\,i},\qquad dx=\frac{t^2-1}{2t^2}\,dt.\] Rationalize the denominators by multiplying the numerator and denominator by the appropriate complex conjugate: \[\frac{1}{2}\left[\frac{t^{i+1}(1-i)}{2} + \frac{t^{i-1}(1+i)}{2}\right] + C =\frac{1}{4}\,t^{i}\left[(1-i)t + (1+i)t^{-1}\right] + C.\]
Substituting back \(t = x - i\sqrt{1-x^2}\) followed by a short algebraic manipulation shows that \[(1-i)t + (1+i)t^{-1} = 2\left(x-\sqrt{1-x^2}\right).\] Then, the final answer becomes \[\int e^{\cos^{-1}(x)}\,dx = \frac{1}{4}\,e^{\cos^{-1}(x)}\cdot 2\left(x-\sqrt{1-x^2}\right)+C = \frac{e^{\cos^{-1}(x)}}{2}\left(x-\sqrt{1-x^2}\right)+C.\]
In USM applications one often meets expressions of the form \[t^n-\frac{1}{t^n}, \qquad n\in\mathbb{N},\] where (as in Transforms 1–2) the parameter \(t\) is chosen piecewise by \[t= \begin{cases} y-\sqrt{y^2-1}, & y\ge 1,\\[6pt] y+\sqrt{y^2-1}, & y\le -1, \end{cases} \qquad\bigl(|y|\ge 1\bigr),\] with principal branches as in the Conventions. (Thus \(t\) is the “small” root on both real components; equivalently, the choice for \(y\le-1\) is the reciprocal of the choice \(y-\sqrt{y^2-1}\).)
Set \[v:=y,\qquad w:=\sqrt{y^2-1},\] so that the two algebraic possibilities are \(v-w\) and \(v+w\). Note that \[(v-w)(v+w)=v^2-w^2=y^2-(y^2-1)=1,\] hence \(\{v-w,\,v+w\}=\{t,\,1/t\}\) regardless of which branch is active.
Define the fixed difference \[D_n(y):=(v-w)^n-(v+w)^n.\] By the binomial theorem, \[(v\pm w)^n=\sum_{k=0}^{n}\binom{n}{k}v^{\,n-k}(\pm w)^k.\] Subtracting the two expansions gives \[D_n(y)=\sum_{k=0}^{n}\binom{n}{k}v^{\,n-k}w^{k}\bigl[(-1)^k-1\bigr] =-2\sum_{\substack{k=0\\ k\;\text{odd}}}^{n}\binom{n}{k}v^{\,n-k}w^k.\] Equivalently, with \(k=2j+1\), \[\boxed{\; D_n(y) =-2\sum_{j=0}^{\lfloor (n-1)/2\rfloor}\binom{n}{2j+1}\, y^{\,n-2j-1}\,\bigl(\sqrt{y^2-1}\bigr)^{2j+1}. \;}\]
Because the USM uses \[t= \begin{cases} v-w, & y\ge 1,\\ v+w, & y\le -1, \end{cases} \qquad\text{and}\qquad \frac{1}{t}= \begin{cases} v+w, & y\ge 1,\\ v-w, & y\le -1, \end{cases}\] we have the domain-dependent relation \[t^n-\frac{1}{t^n} = \begin{cases} D_n(y), & y\ge 1,\\[4pt] -\,D_n(y), & y\le -1. \end{cases}\] Equivalently, \[t^n-\frac{1}{t^n}=\sigma\,D_n(y), \qquad \sigma= \begin{cases} 1,& y\ge 1,\\ -1,& y\le -1. \end{cases}\] Thus, the boxed formula for \(D_n(y)\) may be used uniformly, provided one multiplies by \(\sigma\) to match the branch of \(t\) used in the transforms.
Suppose an antiderivative obtained in \(t\)–space has the form \[I=-\frac{1}{64}\Biggl\{\frac{1}{3}\Bigl[t^3-\frac{1}{t^3}\Bigr]+\Bigl[t^2-\frac{1}{t^2}\Bigr]-\Bigl[t-\frac{1}{t}\Bigr]-4\ln|t|\Biggr\}+C.\] From the boxed identity, \[D_1(y)=-2\sqrt{y^2-1},\qquad D_2(y)=-4y\sqrt{y^2-1},\qquad D_3(y)=-2(4y^2-1)\sqrt{y^2-1}.\] Using \(t^n-t^{-n}=\sigma D_n(y)\) gives \[I=-\frac{1}{64}\left\{\sigma\,\sqrt{y^2-1}\,\Bigl[-\tfrac{2}{3}(4y^2-1)-4y+2\Bigr]-4\ln|t|\right\}+C.\] Finally revert to the original variable using \[y=\frac{x+b}{a},\qquad t= \begin{cases} y-\sqrt{y^2-1} =\dfrac{x+b-\sqrt{(x+b)^2-a^2}}{a}, & y\ge 1,\\[8pt] y+\sqrt{y^2-1} =\dfrac{x+b+\sqrt{(x+b)^2-a^2}}{a}, & y\le -1, \end{cases}\] with principal branches and domain-consistent signs as in the USM transforms.
Let 5 \[Q(x):=a'x^2+b'x+c',\qquad \Delta:=b'^2-4a'c',\qquad X:=x+\frac{b'}{2a'}\quad(a'\ne0).\] Completing the square gives \[Q(x)=a'\!\left(X^2-\frac{\Delta}{4a'^2}\right).\] To avoid clashes with the symbol \(a>0\) used earlier, denote by \[A:=\frac{\sqrt{|\Delta|}}{2|a'|}>0\] the positive radius from the completed square. Depending on the signs of \(a'\) and \(\Delta\), the radical \(\sqrt{Q(x)}\) falls into one of three canonical shapes [8] (the degenerate case \(\Delta=0\) is listed separately): \[\boxed{ \begin{array}{rll} \text{(Difference form)} & \Delta>0,\;a'>0: & \sqrt{Q(x)}=\sqrt{a'}\,\sqrt{X^2-A^2};\\[6pt] \text{(Circular form)} & \Delta>0,\;a'<0: & \sqrt{Q(x)}=\sqrt{|a'|}\,\sqrt{A^2-X^2};\\[6pt] \text{(Sum form)} & \Delta<0,\;a'>0: & \sqrt{Q(x)}=\sqrt{a'}\,\sqrt{X^2+A^2};\\[6pt] \text{(Degenerate)} & \Delta=0: & \sqrt{Q(x)}=\sqrt{|a'|}\,|X|\quad(\text{linear}). \end{array} }\]
In what follows we show that the USM recovers Euler’s first substitution in the difference case via Transform 2, and Euler’s second substitution [8] in the circular case via Transform 5. The sum case is handled directly by the hyperbolic Transform 3 (and matches the usual rational forms after a trivial reparametrization).
Normalize by \[y:=\frac{X}{A},\qquad |y|\ge1,\qquad \alpha=\cos^{-1}(y),\qquad t:=e^{\pm i\alpha}.\] USM Transform 2 yields \[X=A\,\frac{t^2+1}{2t},\qquad S:=\sqrt{X^2-A^2}= \begin{cases} -\,A\,\dfrac{t^2-1}{2t}, & X\ge A\;(y\ge1),\\[8pt] \;\;A\,\dfrac{t^2-1}{2t}, & X\le -A\;(y\le-1), \end{cases}\] which matches the principal real square root \(S\ge0\) on the two real components \((-\infty,-A]\cup[A,\infty)\).
Define the classical Euler combinations \[U_\pm:=X\pm S.\] A short calculation gives the piecewise identities \[(U_+,U_-)= \begin{cases} \big(\dfrac{A}{t},\,A t\big), & X\ge A,\\[8pt] \big(A t,\,\dfrac{A}{t}\big), & X\le -A, \end{cases} \qquad\text{equivalently}\qquad \{U_+,U_-\}=\Big\{A t,\;\dfrac{A}{t}\Big\}.\] Returning to the original quadratic, \[u_\pm:=\sqrt{Q(x)}\;\pm\;\sqrt{a'}\,X=\sqrt{a'}\,(S\pm X),\] so that \[u_+=\sqrt{a'}\,U_+,\qquad u_-=-\,\sqrt{a'}\,U_-.\] Consequently, \[t= \begin{cases} \dfrac{A\sqrt{a'}}{u_+}=\dfrac{u_-}{-A\sqrt{a'}}, & X\ge A,\\[12pt] \dfrac{u_+}{A\sqrt{a'}}=\dfrac{-A\sqrt{a'}}{u_-}, & X\le -A, \end{cases}\] or succinctly: the reconstruction holds up to the trivial reparametrization \(t\leftrightarrow 1/t\) across the two components.
On the two real components \((-\infty,-A]\cup[A,\infty)\), \[t= \begin{cases} \dfrac{X-\sqrt{X^2-A^2}}{A}, & X\ge A,\\[10pt] \dfrac{X+\sqrt{X^2-A^2}}{A}, & X\le -A, \end{cases}\] which is Transform 2 written as a unified real back–substitution with principal branches.
When the leading coefficient is negative and \(\Delta>0\), \[\sqrt{Q(x)}=\sqrt{|a'|}\,\sqrt{A^2-X^2},\qquad |X|\le A.\] Set \[y:=\frac{X}{A}\in[-1,1],\qquad \phi=\sec^{-1}(y),\;\psi=\sin^{-1}(y),\qquad r:=e^{\pm i\phi}=\tan\!\Bigl(\frac{\psi}{2}\Bigr)\] as in USM Transform 5 (cf.Theorem 2). Then \[X=\frac{2A r}{1+r^2},\qquad C:=\sqrt{A^2-X^2}=A\,\frac{1-r^2}{1+r^2}.\] Introduce Euler’s parameter \[t_E:=\frac{C-A}{X}\qquad(\text{note: this is the \emph{negative} of the common textbook choice }(A-C)/X).\] Substituting the USM expressions gives \[t_E=\frac{C-A}{X}=-\,r,\] and hence the reconstructions \[\boxed{\; X=-\,\frac{2A\,t_E}{\,1+t_E^2\,},\qquad C=A\,\frac{1-t_E^2}{\,1+t_E^2\,}. \;}\] At \(X=0\), interpret \(t_E\) by limit: since \[t_E=\frac{C-A}{X}=-\frac{X}{C+A},\] we have \(t_E\to 0\) as \(X\to 0\), with \(t_E\to 0^{-}\) as \(X\to 0^{+}\) and \(t_E\to 0^{+}\) as \(X\to 0^{-}\).
Returning to \(Q\), \[\boxed{\; \sqrt{Q(x)}=\sqrt{|a'|}\,C=\sqrt{|a'|}\!\left(A+\Bigl(x+\frac{b'}{2a'}\Bigr)t_E\right),\qquad x=-\frac{b'}{2a'}-\frac{2A\,t_E}{1+t_E^2}. \;}\] Thus the circular USM reparametrization is literally Euler’s second substitution with \(r=-t_E\).
When \(\Delta<0\) and \(a'>0\), \[\sqrt{Q(x)}=\sqrt{a'}\,\sqrt{X^2+A^2}.\] With \[y:=\frac{X}{A},\qquad \theta=\sinh^{-1}(y),\qquad s:=e^{\theta}=y+\sqrt{y^2+1},\] USM Transform 3 gives \[X=A\,\frac{s^2-1}{2s},\qquad H:=\sqrt{X^2+A^2}=A\,\frac{s^2+1}{2s},\qquad dx=A\,\frac{s^2+1}{2s^2}\,ds.\] This handles the sum form uniformly. After the reparametrization \(t=1/s\) one recovers the familiar rational expressions used in Euler-type treatments (up to harmless overall sign conventions).
Here \(Q(x)=a'X^2\) and \(\sqrt{Q(x)}=\sqrt{|a'|}\,|X|\) is linear after the shift \(X\). No special substitution is required.
This note develops a unified substitution method (USM) for integrals involving quadratic radicals and half-angle inverse–trigonometric composites by centering the analysis on exponentials of principal inverse functions. The key observation is that, on principal branches, the quantities \(e^{\pm i\cos^{-1}(\cdot)}\) and \(e^{\pm i\sec^{-1}(\cdot)}\) admit simple algebraic representatives that encode the familiar “small root” choices automatically. From these two identities (together with a short bridge lemma linking the circular and hyperbolic regimes), we derived five explicit transforms that map broad families of circular and hyperbolic integrands to rational functions of a single parameter, with Jacobians that are independent of the \(\pm\) sign once the branch is fixed.
The framework consolidates several classical techniques under one branch-consistent calculus. Euler’s first and second substitutions appear as direct instances of Transforms 2 and 5 up to trivial reparametrizations, and the Weierstrass half-angle substitution arises cleanly as a unit-radius specialization of Transform 5. In addition, the binomial–difference identity provides a compact mechanism for simplifying ubiquitous back-substitution patterns such as \(t^{n}-t^{-n}\), reducing algebraic overhead in the final expressions. The worked examples illustrate how the transforms often “pre-cancel” radical factors against the Jacobian, turning what would otherwise be multi-stage trigonometric/hyperbolic workflows into a single rational step; the preliminary CAS benchmark reported in the text suggests this can translate into improved predictability and reduced expression swell for structurally mixed inputs.
These identities originated in efforts to generalize Mollweide’s formulas [1], which serendipitously led us to discover an entire family of trigonometric–hyperbolic formulas for the roots of quadratic equations [2].↩︎
The name Weierstrass substitution is conventional rather than historical. The tangent half-angle parameterization \(t=\tan(\omega/2)\), which rationalizes trigonometric expressions, appears in Euler’s Institutionum calculi integralis (1768) and is treated systematically in Legendre’s Exercices de calcul intégral (1817); see [4], [5]. A concise discussion of the resulting misattribution is given by Johnson [6]. For a standard modern reference using the conventional terminology, see [7].↩︎
This integral comes from the 2006 MIT Integration Bee; none of the competitors managed to solve it within the allotted time. See Keith Winstein, MIT 2006 Integration Bee, YouTube video, posted 7 July 2010, https://www.youtube.com/watch?v=qQ-56b_LvOw&t=3877 (accessed 22 January 2025), at 1:04:38.↩︎
We performed a comprehensive benchmark on a dataset of 100 integrals to compare the USM against Mathematica’s Integrate. The dataset consists of integrands systematically constructed from \(x^k\), \(\tan(\frac{1}{2}\csc^{-1}(\frac{x+b}{a}))\), and \(\tan(\frac{1}{2}\sec^{-1}(\frac{x+b}{a}))\), including rational combinations and products thereof.
The USM approach utilized the “Transform 1” paired with the piecewise back-substitution (summing the computation times for the \(y \ge 1\) and \(y \le -1\) branches). Results indicate that
USM is faster than Integrate in 82 out of 100 cases. Specifically, USM exhibits predictable performance and achieves order-of-magnitude speedups on structurally complex mixed integrands where Integrate often exceeds 1–3 seconds.
While Integrate is faster on simple, pattern-friendly cases, it exhibits higher variance and occasional result bloating: “monster” antiderivatives (ByteCnt \(\ge\) 10,000) occurred 24 times with
Integrate versus only 5 times with USM. Full benchmark tables and code are available in [3].↩︎
Leonhard Euler himself introduced only the first two substitutions that now bear his name; the familiar “third Euler substitution” seems to have been added later by subsequent authors. See J. L. Cieśliński and M. Jurgielewicz, On geometric interpretation of Euler’s substitutions, arXiv:2310.12160 (2023), § 2.4.↩︎