View source for Integral of the secant function

{{Short description|Antiderivative of the secant function}}
[[File:Graph of the integral of the secant.png|thumb|right|upright=1.3|A graph of the secant function (red) and its antiderivative (blue)]]
{{calculus|expanded=integral}}
In [[calculus]], the '''integral of the secant function''' can be evaluated using a variety of methods and there are multiple ways of expressing the [[antiderivative]], all of which can be shown to be equivalent via [[trigonometric identity|trigonometric identities]],

:<math>\int \sec \theta \, d\theta = \begin{cases}
\dfrac12 \ln \dfrac{1+\sin\theta}{1-\sin\theta} + C \\[15mu]
\ln{\bigl|\sec\theta + \tan\theta\,\bigr|} + C \\[15mu]
\ln{\left|\,{\tan}\biggl(\dfrac\theta2 + \dfrac\pi4\biggr) \right|} + C
\end{cases}</math>
This formula is useful for evaluating various [[trigonometric integral]]s. In particular, it can be used to evaluate the [[Integral of secant cubed|integral of the secant  cubed]], which, though seemingly special, comes up rather frequently in applications.<ref name=stewart/>

The definite integral of the secant function starting from <math>0</math> is the inverse [[Gudermannian function]], <math display=inline>\operatorname{gd}^{-1}.</math> For numerical applications, all of the above expressions result in [[loss of significance]] for some arguments. An alternative expression in terms of the [[inverse hyperbolic sine]] {{math|arsinh}} is numerically well behaved for real arguments {{nobr|<math display=inline>|\phi| < \tfrac12\pi</math>:<ref>For example this form is used in {{cite journal |last=Karney |first=Charles F.F. |year=2011 |title=Transverse Mercator with an accuracy of a few nanometers. |journal=Journal of Geodesy |volume=85 |pages=475-485 }}</ref>}}

:<math>
\operatorname{gd}^{-1}\phi = \int_0^\phi \sec \theta \, d\theta
= \operatorname{arsinh}(\tan \phi).
</math>

The integral of the secant function was historically one of the first integrals of its type ever evaluated, before most of the development of integral calculus. It is important because it is the vertical coordinate of the [[Mercator projection]], used for [[rhumb line|marine navigation with constant compass bearing]].

== Proof that the different antiderivatives are equivalent ==

=== Trigonometric forms ===
Three common expressions for the integral of the secant,
:<math>\begin{align}
\int \sec \theta \, d\theta
&= \dfrac12 \ln \dfrac{1+\sin\theta}{1-\sin\theta} + C \\[5mu]
&= \ln{\bigl|\sec\theta + \tan\theta\,\bigr|} + C \\[5mu]
&= \ln{\left|\,{\tan}\biggl(\frac\theta2 + \frac\pi4\biggr) \right|} + C,
\end{align}</math>

are equivalent because

:<math>
\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}
= \bigl|\sec\theta + \tan\theta\,\bigr|
= \left|\,{\tan}\biggl(\frac\theta2 + \frac\pi4\biggr) \right|.
</math>

Proof: we can separately apply the [[tangent half-angle substitution]] <math>t = \tan\tfrac12 \theta</math> to each of the three forms, and show them equivalent to the same expression in terms of <math>t.</math> Under this substitution <math>\cos \theta = (1 - t^2)\big/(1 + t^2)</math> and <math>\sin \theta = 2t \big/(1 + t^2).</math>

First, 

:<math>\begin{align}
\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}
&= \sqrt{\frac{1 + \dfrac{2t}{1 + t^2}}{1 - \dfrac{2t}{1 + t^2}}}
= \sqrt{\frac{1 + t^2 + 2t}{1 + t^2 - 2t}}
= \sqrt{\frac{(1 + t)^2}{(1 - t)^2}} \\[5mu]
&= \left|\frac{1 + t}{1 - t}\right|.
\end{align}</math>

Second,

:<math>\begin{align}
\bigl|\sec\theta + \tan\theta\,\bigr|
&= \left|\frac1{\cos\theta} + \frac{\sin\theta}{\cos\theta}\right|
= \left|\frac{1 + t^2}{1 - t^2} + \frac{2t}{1 - t^2}\right|
= \left|\frac{(1 + t)^2}{(1 + t)(1 - t)}\right| \\[5mu]
&= \left|\frac{1 + t}{1 - t}\right|.
\end{align}</math>

Third, using the tangent addition identity <math>\tan(\phi + \psi) = (\tan\phi + \tan\psi) \big/ (1 - \tan\phi\,\tan\psi),</math>

:<math>\begin{align}
\left|\,{\tan}\biggl(\frac\theta2 + \frac\pi4\biggr) \right|
&= \left|\frac{\tan\tfrac12 \theta + \tan\tfrac14\pi}{1 - \tan\tfrac12 \theta \, \tan\tfrac14\pi} \right|
= \left|\frac{t + 1}{1 - t \cdot 1}\right| \\[5mu]
&= \left|\frac{1 + t}{1 - t}\right|.
\end{align}</math>

So all three expressions describe the same quantity.

The conventional solution for the [[Mercator projection]] ordinate may be written without the [[absolute value]] signs since the latitude <math>\varphi</math> lies between <math display=inline>-\tfrac12\pi</math> and <math display=inline>\tfrac12\pi</math>,
:<math>y = \ln\,{\tan}\biggl(\frac\varphi2 + \frac\pi4\biggr).</math>

=== Hyperbolic forms ===
Let
:<math>
\begin{align}
  \psi      &=\ln(\sec\theta+\tan\theta),\\[4pt]
  e^\psi    &=\sec\theta+\tan\theta,\\[4pt]
\sinh\psi   &=\frac{e^\psi-e^{-\psi}}{2}=\tan\theta,\\[4pt]
\cosh\psi   &=\sqrt{1+\sinh^2\psi}= |\sec\theta\, |,\\[4pt]
\tanh\psi   &=\sin\theta.
  \end{align}
</math>
Therefore,
:<math>\begin{align}
  \int \sec \theta \, d\theta 
    &=\operatorname{artanh} \left(\sin\theta\right) + C \\[-2mu]
    &=\sgn(\cos \theta)\operatorname{arsinh} \left( \tan \theta\right)+C \\[7mu]  
    &=\sgn(\sin \theta) \operatorname{arcosh}{ \left| \sec \theta\right|}+C.
\end{align}</math>

== History ==
{{Broader|Mercator projection#History}}
The [[integral]] of the [[secant function]] was one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 by [[James Gregory (mathematician)|James Gregory]].<ref name="rickey-tuchinsky"/> He applied his result to a problem concerning nautical tables.<ref name=stewart/> In 1599, [[Edward Wright (mathematician)|Edward Wright]] evaluated the integral by [[numerical method]]s – what today we would call [[Riemann sum]]s.<ref name=wright/> He wanted the solution for the purposes of [[cartography]] – specifically for constructing an accurate [[Mercator projection]].<ref name="rickey-tuchinsky"/> In the 1640s, Henry Bond, a teacher of navigation, surveying, and other mathematical topics, compared Wright's numerically computed table of values of the integral of the secant with a table of logarithms of the tangent function, and consequently [[conjecture]]d that<ref name="rickey-tuchinsky"/>

:<math>\int_0^\varphi \sec\theta\,d\theta = \ln\tan\left(\frac{\varphi}{2} + \frac{\pi}{4}\right).</math>

This conjecture became widely known, and in 1665, [[Isaac Newton]] was aware of it.<ref name=newton/>

==Evaluations==
=== By a standard substitution (Gregory's approach) ===
A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator by {{math|sec&thinsp;''θ'' + tan&thinsp;''θ''}} and then using the substitution {{math|''u'' {{=}} sec&thinsp;''θ'' + tan&thinsp;''θ''}}. This substitution can be obtained from the [[derivative]]s of secant and tangent added together, which have secant as a common factor.<ref name=handouts/>

Starting with

:<math>\frac d{d\theta}\sec \theta = \sec\theta \tan\theta \quad \text{and} \quad \frac d{d\theta}\tan \theta = \sec^2 \theta,</math>

adding them gives

:<math>\begin{align}
\frac d{d\theta}(\sec\theta + \tan\theta) &= \sec\theta \tan\theta + \sec^2\theta \\
&= \sec\theta (\tan\theta + \sec \theta).
\end{align}</math>

The derivative of the sum is thus equal to the sum multiplied by {{math|sec&thinsp;''θ''}}. This enables multiplying {{math|sec&thinsp;''θ''}} by {{math|sec&thinsp;''θ'' + tan&thinsp;''θ''}} in the numerator and denominator and performing the following substitutions:

:<math>\begin{align}
u &= \sec \theta + \tan \theta \\
du &= \left(\sec \theta \tan \theta + \sec^2 \theta\right)\,d\theta.
\end{align}</math>

The integral is evaluated as follows:

: <math>\begin{align}
\int \sec \theta \,d\theta &= \int \frac{\sec\theta (\sec\theta + \tan\theta)}{\sec\theta + \tan\theta} \,d\theta \\[6pt]
&= \int \frac{\sec^2\theta + \sec\theta \tan\theta}{\sec\theta + \tan\theta}\,d\theta & u &= \sec\theta + \tan\theta \\[6pt]
&= \int \frac{1}{u}\,du & du &= \left(\sec\theta \tan\theta + \sec^2\theta\right) \,d\theta \\[6pt]
&= \ln |u| + C \\[4pt]
&= \ln |\sec\theta + \tan\theta| + C,
\end{align}</math>

as claimed. This was the formula discovered by James Gregory.<ref name=stewart/>

===By partial fractions and a substitution (Barrow's approach)===
Although Gregory [[mathematical proof|proved]] the conjecture in 1668 in his ''Exercitationes Geometricae'',<ref name=gregory/> the proof was presented in a form that renders it nearly impossible for modern readers to comprehend; [[Isaac Barrow]], in his ''Lectiones Geometricae'' of 1670,<ref name=barrow/> gave the first "intelligible" proof, though even that was "couched in the geometric idiom of the day."<ref name="rickey-tuchinsky"/> Barrow's proof of the result was the earliest use of [[partial fractions]] in integration.<ref name="rickey-tuchinsky"/>  Adapted to modern notation, Barrow's proof began as follows:

:<math>
\int \sec \theta \, d\theta
= \int \frac{1}{\cos\theta} \,d\theta
= \int \frac{\cos\theta }{\cos^2\theta}\, d\theta
= \int \frac{\cos\theta }{1 - \sin^2\theta} \, d\theta
</math>

Substituting {{math|''u'' {{=}} sin&thinsp;''θ''}}, {{math|''du'' {{=}} cos&thinsp;''θ'' ''dθ''}}, reduces the integral to

:<math>\begin{align}
\int \frac{1}{1 - u^2} \,du &= \int\frac{1}{(1+u)(1-u)} \,du \\[6pt]
&= \int \frac12 \!\left(\frac{1}{1+u} + \frac{1}{1-u}\right) du &&\text{partial fraction decomposition} \\[6pt]
&= \frac12 \bigl( \ln\left|1 + u\right|-\ln \left|1-u\right| \bigr) + C \\[6pt]
&= \frac12 \ln\left|\frac{1+u}{1-u}\right| + C
\end{align}</math>

Therefore,

:<math>
\int \sec \theta \,d\theta
= \frac{1}{2}\ln \frac{1 + \sin \theta}{1 - \sin \theta} + C,
</math>

as expected. Taking the absolute value is not necessary because <math>1 + \sin \theta</math> and <math>1 - \sin \theta</math> are always non-negative for real values of <math>\theta.</math>

===By the tangent half-angle substitution ===

==== Standard ====
Under the [[tangent half-angle substitution]] <math display=inline>t = \tan\tfrac12\theta,</math><ref name=stewart2/>

:<math>\begin{align}
&\sin \theta = \frac{2t}{1 + t^2},\quad
\cos \theta = \frac{1 - t^2}{1 + t^2},\quad
d\theta = \frac{2}{1 + t^2}\,dt,
\\[10mu]
&\tan \theta = \frac{\sin\theta}{\cos\theta} = \frac{2t}{1 - t^2},\quad
\sec \theta = \frac1{\cos\theta} = \frac{1 + t^2}{1 - t^2},
\\[10mu]
&\sec \theta + \tan \theta = \frac{1 + 2t + t^2}{1 - t^2} = \frac{1 + t}{1 - t}.
\end{align}</math>

Therefore the integral of the secant function is

:<math>\begin{align}
\int \sec \theta \,d\theta
&= \int \left( \frac{1 + t^2}{1 - t^2} \right)\!\left( \frac{2}{1 + t^2} \right) dt && t = \tan\frac{\theta}{2}\\[6pt]
&= \int \frac{2}{(1 - t) (1 + t)} \, dt \\[6pt]
&= \int \left( \frac{1}{1+t} + \frac{1}{1-t} \right) dt && \text{partial fraction decomposition} \\[6pt]
&= \ln|1+t| - \ln|1-t| + C \\[6pt]
&= \ln \left| \frac{1+t}{1-t}\right| + C\\[6pt]
&= \ln | \sec \theta + \tan \theta| + C,
\end{align}</math>

as before.

==== Non-standard ====
The integral can also be derived by using a somewhat non-standard version of the tangent half-angle substitution, which is simpler in the case of this particular integral, published in 2013,<ref name=hardy/> is as follows:

: <math>\begin{align}
x &= \tan \left( \frac \pi 4 + \frac \theta 2 \right) \\[10pt]
\frac{2x}{1+x^2} &= \frac{2 \tan \left( \frac \pi 4 + \frac \theta 2 \right)}{\sec^2 \left( \frac \pi 4 + \frac \theta 2 \right)} = 2\sin \left( \frac \pi 4 + \frac \theta 2 \right) \cos \left( \frac \pi 4 + \frac \theta 2 \right)\\[6pt]
&=\sin \left(\frac{\pi}{2}+\theta \right)=\cos\theta &&\text{by the double-angle formula}\\[10pt]
dx&=\frac12 \sec^2 \left(\frac{\pi}{4}+\frac{\theta}{2}\right)d\theta=\frac12 \left(1+x^2\right) d\theta\\[10pt]
d\theta &=\frac{2}{1+x^2}\,dx.
\end{align}</math>
Substituting:
: <math>\begin{align}
\int \sec\theta \, d\theta = \int \frac{1}{\cos\theta} \, d\theta & = \int \frac{1+x^2}{2x} \cdot \frac{2}{1+x^2} \,dx \\[6pt]
&= \int \frac{1}{x} \,dx \\[6pt]
&= \ln|x| + C \\[6pt]
&= \ln\left| \tan\left( \frac \pi 4 + \frac \theta 2 \right) \right| + C.
\end{align}</math>

=== By two successive substitutions ===

The integral can also be solved by manipulating the [[integrand]] and substituting twice. Using the definition {{math|sec&thinsp;''θ'' {{=}} {{sfrac|1|cos&thinsp;''θ''}}}} and the identity {{math|cos<sup>2</sup>&thinsp;''θ'' + sin<sup>2</sup>&thinsp;''θ'' {{=}} 1}}, the integral can be rewritten as

: <math>\int \sec \theta \, d\theta=\int \frac{1}{\cos \theta}\,d\theta = \int \frac{\cos \theta}{\cos^2 \theta}\,d\theta=\int \frac{\cos \theta}{1-\sin^2 \theta}\,d\theta.</math>

Substituting {{math|''u'' {{=}} sin&thinsp;''θ''}}, {{math|''du'' {{=}} cos&thinsp;''θ'' ''dθ''}} reduces the integral to

: <math>\int \frac{1}{1-u^2} \, du.</math>

The reduced integral can be evaluated by substituting {{math|''u'' {{=}} tanh&thinsp;''t''}}, {{math|''du'' {{=}} sech<sup>2</sup>&thinsp;''t'' ''dt''}}, and then using the identity {{math|1 − tanh<sup>2</sup>&thinsp;''t'' {{=}} sech<sup>2</sup>&thinsp;''t''}}.

: <math>\int \frac{\operatorname{sech}^2 t}{1-\tanh^2 t}\,dt = \int \frac{\operatorname{sech}^2 t}{\operatorname{sech}^2 t} \, dt = \int dt.</math>

The integral is now reduced to a simple integral, and back-substituting gives

:<math>\begin{align}
\int dt &= t+C \\
&=\operatorname{artanh} u+C \\[4pt]
&=\operatorname{artanh}(\sin \theta)+C,
\end{align}</math>

which is one of the hyperbolic forms of the integral.

A similar strategy can be used to integrate the [[cosecant]], [[hyperbolic secant]], and [[hyperbolic cosecant]] functions.

=== Other hyperbolic forms ===

It is also possible to find the other two hyperbolic forms directly, by again multiplying and dividing by a convenient term:

:<math>\int \sec \theta \,d\theta=\int \frac{\sec^2 \theta}{\sec \theta} \,d\theta=\int \frac{\sec^2 \theta}{\pm\sqrt{1+\tan^2 \theta}} \,d\theta,</math>

where <math>\pm</math> stands for <math>\sgn(\cos \theta)</math> because  <math>\sqrt{1+\tan^2 \theta} = |\sec\theta\,|.</math> Substituting {{math|''u'' {{=}} tan&thinsp;''θ''}}, {{math|''du'' {{=}} sec<sup>2</sup>&thinsp;''θ'' ''dθ''}}, reduces to a standard integral:

:<math>\begin{align}
\int \frac{1}{\pm\sqrt{1+u^2}} \,du
&=\pm\operatorname{arsinh} u+C \\
&=\sgn(\cos \theta)\operatorname{arsinh} \left( \tan \theta\right)+C,
\end{align}</math>

where {{math|sgn}} is the [[sign function]].

Likewise:

:<math>\int \sec \theta \,d\theta =\int \frac{\sec \theta \tan \theta}{\tan \theta} \,d\theta=\int \frac{\sec \theta \tan \theta}{\pm\sqrt{\sec^2 \theta-1}} \,d\theta.</math>

Substituting {{math|''u'' {{=}} {{abs|sec&thinsp;''θ''}}}}, {{math|''du'' {{=}} {{abs|sec&thinsp;''θ''}} tan&thinsp;''θ'' ''dθ''}}, reduces to a standard integral:

:<math>\begin{align}
\int \frac{1}{\pm\sqrt{u^2-1}} \,du &= \pm\operatorname{arcosh} u+C \\
&=\sgn(\sin \theta) \operatorname{arcosh} \left| \sec \theta\right|+C.
\end{align}</math>

=== Using complex exponential form ===

Under the substitution <math>z = e^{i\theta},</math>

:<math>\begin{align}
&\theta = -i \ln z,\quad
d\theta = \frac{-i}z dz,\quad
\cos \theta = \frac{z + z^{-1}}2,\quad
\sin \theta = \frac{z - z^{-1}}{2i},\quad \\[5mu]
&\sec \theta = \frac2{z + z^{-1}},\quad
\tan \theta = -i\frac{z - z^{-1}}{z + z^{-1}},\quad \\[5mu]
&\sec \theta + \tan \theta = -i\frac{2i + z - z^{-1}}{z + z^{-1}}
= -i\frac{(z + i)(1 + iz^{-1})}{(z - i)(1 + iz^{-1})}
= -i\frac{z + i}{z - i}
\end{align}</math>

So the integral can be solved as:
:<math>\begin{align}
\int \sec \theta \,d\theta
&= \int \frac2{z + z^{-1}}\, \frac{-i}z dz && z = e^{i\theta} \\[5mu]
&= \int \frac{-2i}{z^2 + 1} dz \\
&= \int \frac{1}{z+i} - \frac{1}{z-i} \,dz && \text{partial fraction decomposition} \\[5mu]
&= \ln(z+i)-\ln(z-i) + C \\[5mu]
&= \ln\frac{z + i}{z - i} + C \\[5mu]
&= \ln\bigl(i(\sec \theta + \tan \theta)\bigr)  + C \\[5mu]
&= \ln(\sec \theta + \tan \theta) + \ln i + C
\end{align}</math>

Because the constant of integration can be anything, the additional constant term can be absorbed into it. Finally, if theta is [[real number|real]]-valued, we can indicate this with absolute value brackets in order to get the equation into its most familiar form:
:<math>
\int \sec \theta \,d\theta = \ln\left|\tan\theta + \sec\theta\right| + C
</math>

==Gudermannian and Lambertian==
[[File:Gudermannian function.png|thumb|upright=1.5|The Gudermannian function relates the [[area]] of a [[circular sector]] to the area of a [[hyperbolic sector]], via a common [[stereographic projection]]. If twice the area of the blue hyperbolic sector is {{math|''ψ''}}, then twice the area of the red circular sector is {{math|''ϕ'' {{=}} gd ''ψ''}}. Twice the area of the purple triangle is the stereographic projection {{math|''s'' {{=}} tan&thinsp;{{sfrac|1|2}}''ϕ'' {{=}} tanh&thinsp;{{sfrac|1|2}}''ψ''.}} The blue point has coordinates {{math|(cosh&thinsp;''ψ'', sinh&thinsp;''ψ'')}}. The red point has coordinates {{math|(cos&thinsp;''ϕ'', sin&thinsp;''ϕ'').}} The purple point has coordinates {{math|(0, ''s'').}}  ]]

The integral of the hyperbolic secant function defines the [[Gudermannian function]]:
:<math>\int_0^\psi \operatorname{sech} u \, du
=\operatorname{gd}\psi.</math>

The integral of the secant function defines the Lambertian function, which is the [[inverse function|inverse]] of the Gudermannian function:
:<math>\int_0^\varphi \sec t \, dt
=\operatorname{lam}\varphi
=\operatorname{gd}^{-1}\varphi.</math>

These functions are encountered in the theory of map projections: the [[Mercator projection]] of a point on the [[sphere]] with longitude {{mvar|λ}} and latitude {{mvar|ϕ}} may be written<ref name=lee/> as: 

:<math>(x,y) = \bigl(\lambda,\operatorname{lam}\varphi\bigr).</math>

== See also ==
{{Portal|Mathematics}}
*[[Integral of secant cubed]]

{{clear}}

== Notes ==
{{reflist |30em |refs=

<ref name=gregory>
{{cite book
| last = Gregory
| first = James
| author-link = James Gregory (mathematician)
| year = 1668
| title = Exercitationes Geometricae
| trans-title = Geometrical Exercises
| language = la
| publisher = Moses Pitt
| chapter = Analogia Inter Lineam Meridianam Planispherii Nautici & Tangentes Artificiales Geometricè Demonstrata, &c. 
| trans-chapter = Analogy Between the Meridian Line of the Nautical Planisphere & Artificial Tangents Geometrically Demonstrated, &c.
| pages = 14–24
| chapter-url = https://archive.org/details/ita-bnc-mag-00001357-002/page/n27/
}}
</ref>

<ref name=stewart>
{{Cite book
| last = Stewart
| first = James
| author-link = James Stewart (mathematician)
| year = 2012
| title = Calculus - Early Transcendentals
| publisher = Cengage Learning
| isbn = 978-0-538-49790-9
| pages = 475-6
| chapter = Section 7.2: Trigonometric Integrals
}}
</ref>

<ref name=stewart2>
{{Cite book
| last = Stewart
| first = James
| title = Calculus: Early Transcendentals
| url = https://archive.org/details/singlevariableca00stew_704
| url-access = limited
| publisher = Cengage Learning
| year = 2012
| isbn = 978-0-538-49790-9
| edition = 7th
| location = Belmont, CA, USA
| pages = [https://archive.org/details/singlevariableca00stew_704/page/n524 493]
| chapter = Section 7.4: Integration of Rational Functions by Partial Fractions
}}
</ref>

<ref name="rickey-tuchinsky">V. Frederick Rickey and Philip M. Tuchinsky, [https://www.jstor.org/stable/2690106 An Application of Geography to Mathematics: History of the Integral of the Secant] in  [[Mathematics Magazine]], volume 53, number 3, May 1980, pages 162–166.
</ref>

<ref name=wright>[[Edward Wright (mathematician)|Edward Wright]], ''Certaine Errors in Navigation, Arising either of the ordinaire erroneous making or vsing of the sea Chart, Compasse, Crosse staffe, and Tables of declination of the Sunne, and fixed Starres detected and corrected'', Valentine Simms, London, 1599.</ref>

<ref name=newton>H. W. Turnbull, editor, ''The Correspondence of Isaac Newton'', Cambridge University Press, 1959–1960, volume 1, pages 13–16 and volume 2, pages 99–100.
<p>[[D. T. Whiteside]], editor, ''The Mathematical Papers of Isaac Newton'', Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.</p>
</ref>

<ref name=handouts>
{{cite web
| last = Feldman
| first = Joel
| title = Integration of sec x and sec<sup>3</sup> x
| url = https://www.math.ubc.ca/~feldman/m121/secx.pdf
| website = University of British Columbia Mathematics Department
}}
<p>{{cite web 
| title = Integral of Secant
| url = https://ocw.mit.edu/courses/18-01sc-single-variable-calculus-fall-2010/resources/mit18_01scf10_ses71c/
| website = MIT OpenCourseWare
}}</p>
</ref>

<ref name=barrow>
{{cite book
| last = Barrow
| first = Isaac
| authorlink = Isaac Barrow
| year = 1674 
| title = Lectiones Opticae & Geometricae
| lang = la
| publisher = Typis Guilielmi Godbid
| orig-year=1670 
| chapter = Lectiones geometricae: XII, Appendicula I
| pages = 110–114
| chapter-url = https://archive.org/details/lectionesoptic00barr/page/110/
}} In English, {{cite book
| last = Barrow
| first = Isaac
| display-authors = 0
| translator-last = Child
| translator-first = James Mark
| year = 1916
| publisher = Open Court
| title = The Geometrical Lectures of Isaac Barrow
| chapter = Lecture XII, Appendix I
| chapter-url = https://archive.org/details/geometricallectu00barr/page/165
| pages = 165–169
}}
</ref>

<ref name=hardy>
{{cite journal
| last = Hardy
| first = Michael 
| year = 2013
| title = Efficiency in Antidifferentiation of the Secant Function
| journal = American Mathematical Monthly
| volume = 120
| number = 6
| page = 580
| url = https://www.tandfonline.com/doi/abs/10.4169/amer.math.monthly.120.06.580
| url-access = subscription
}}
</ref> <!-- the DOI for this paper is broken -->

<ref name=lee>
{{cite book
| last = Lee | first = L. P. | author-link = Laurence Patrick Lee
| year = 1976
| title = Conformal Projections Based on Elliptic Functions
| location = Toronto | publisher = B. V. Gutsell, York University
| series = ''Cartographica Monographs'' | volume = 16
| url = https://archive.org/details/conformalproject0000leel | url-access = limited
| isbn = 0-919870-16-3
}} Supplement No. 1 to [https://www.utpjournals.press/toc/cart/13/1 ''The Canadian Cartographer'' '''13'''].
</ref>

}} <!-- end footnote definitions -->

== References ==

* {{cite book |editor-last=Maseres |editor-first=Francis |year=1791 |volume=2 |title=Scriptores Logarithmici; Or, A Collection of Several Curious Tracts on the Nature and Construction of Logarithms |publisher=J. Davis |url=https://archive.org/details/scriptoreslogari02mase/ }}
* {{cite journal |last=Strauss |first=Simon W. |year=1980 |title=The Integrals <math display=inline>\int\sec x\,dx</math> and <math display=inline>\int\csc x\,dx</math> Revisited |journal=Journal of the Washington Academy of Sciences |volume=70 |number=4 |pages=137–143 |jstor=24537231 }}

{{Calculus topics}}

[[Category:Integral calculus]]