Informatics Educational Institutions & Programs

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 identities,

\int \sec \theta \,d\theta ={\begin{cases}{\dfrac {1}{2}}\ln {\dfrac {1+\sin \theta }{1-\sin \theta }}+C\\[15mu]\ln {{\bigl |}\sec \theta +\tan \theta \,{\bigr |}}+C\\[15mu]\ln {\left|\,{\tan }{\biggl (}{\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}{\biggr )}\right|}+C\end{cases}}

This formula is useful for evaluating various trigonometric integrals. In particular, it can be used to evaluate the integral of the secant cubed, which, though seemingly special, comes up rather frequently in applications.^[1]

The definite integral of the secant function starting from $0$ is the inverse Gudermannian function, ${\textstyle \operatorname {gd} ^{-1}.}$ 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 $arsinh$ is numerically well behaved for real arguments ${\textstyle |\phi |<{\tfrac {1}{2}}\pi }$ :^[2]

\operatorname {gd} ^{-1}\phi =\int _{0}^{\phi }\sec \theta \,d\theta =\operatorname {arsinh} (\tan \phi ).

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 marine navigation with constant compass bearing.

Proof that the different antiderivatives are equivalent

Trigonometric forms

Three common expressions for the integral of the secant,

{\begin{aligned}\int \sec \theta \,d\theta &={\dfrac {1}{2}}\ln {\dfrac {1+\sin \theta }{1-\sin \theta }}+C\\[5mu]&=\ln {{\bigl |}\sec \theta +\tan \theta \,{\bigr |}}+C\\[5mu]&=\ln {\left|\,{\tan }{\biggl (}{\frac {\theta }{2}}+{\frac {\pi }{4}}{\biggr )}\right|}+C,\end{aligned}}

are equivalent because

{\sqrt {\dfrac {1+\sin \theta }{1-\sin \theta }}}={\bigl |}\sec \theta +\tan \theta \,{\bigr |}=\left|\,{\tan }{\biggl (}{\frac {\theta }{2}}+{\frac {\pi }{4}}{\biggr )}\right|.

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

First,

{\begin{aligned}{\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{aligned}}

Second,

{\begin{aligned}{\bigl |}\sec \theta +\tan \theta \,{\bigr |}&=\left|{\frac {1}{\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{aligned}}

Third, using the tangent addition identity $\tan(\phi +\psi )=(\tan \phi +\tan \psi ){\big /}(1-\tan \phi \,\tan \psi ),$

{\begin{aligned}\left|\,{\tan }{\biggl (}{\frac {\theta }{2}}+{\frac {\pi }{4}}{\biggr )}\right|&=\left|{\frac {\tan {\tfrac {1}{2}}\theta +\tan {\tfrac {1}{4}}\pi }{1-\tan {\tfrac {1}{2}}\theta \,\tan {\tfrac {1}{4}}\pi }}\right|=\left|{\frac {t+1}{1-t\cdot 1}}\right|\\[5mu]&=\left|{\frac {1+t}{1-t}}\right|.\end{aligned}}

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 $\varphi$ lies between ${\textstyle -{\tfrac {1}{2}}\pi }$ and ${\textstyle {\tfrac {1}{2}}\pi }$ ,

y=\ln \,{\tan }{\biggl (}{\frac {\varphi }{2}}+{\frac {\pi }{4}}{\biggr )}.

Hyperbolic forms

Let

{\begin{aligned}\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{aligned}}

Therefore,

{\begin{aligned}\int \sec \theta \,d\theta &=\operatorname {artanh} \left(\sin \theta \right)+C\\[-2mu]&=\operatorname {sgn}(\cos \theta )\operatorname {arsinh} \left(\tan \theta \right)+C\\[7mu]&=\operatorname {sgn}(\sin \theta )\operatorname {arcosh} {\left|\sec \theta \right|}+C.\end{aligned}}

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.^[3] He applied his result to a problem concerning nautical tables.^[1] In 1599, Edward Wright evaluated the integral by numerical methods – what today we would call Riemann sums.^[4] He wanted the solution for the purposes of cartography – specifically for constructing an accurate Mercator projection.^[3] 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 conjectured that^[3]

\int _{0}^{\varphi }\sec \theta \,d\theta =\ln \tan \left({\frac {\varphi }{2}}+{\frac {\pi }{4}}\right).

This conjecture became widely known, and in 1665, Isaac Newton was aware of it.^[5]

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 $sec θ + tan θ$ and then using the substitution $u = sec θ + tan θ$ . This substitution can be obtained from the derivatives of secant and tangent added together, which have secant as a common factor.^[6]

Starting with

{\frac {d}{d\theta }}\sec \theta =\sec \theta \tan \theta \quad {\text{and}}\quad {\frac {d}{d\theta }}\tan \theta =\sec ^{2}\theta ,

adding them gives

{\begin{aligned}{\frac {d}{d\theta }}(\sec \theta +\tan \theta )&=\sec \theta \tan \theta +\sec ^{2}\theta \\&=\sec \theta (\tan \theta +\sec \theta ).\end{aligned}}

The derivative of the sum is thus equal to the sum multiplied by $sec θ$ . This enables multiplying $sec θ$ by $sec θ + tan θ$ in the numerator and denominator and performing the following substitutions:

{\begin{aligned}u&=\sec \theta +\tan \theta \\du&=\left(\sec \theta \tan \theta +\sec ^{2}\theta \right)\,d\theta .\end{aligned}}

The integral is evaluated as follows:

{\begin{aligned}\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{aligned}}

as claimed. This was the formula discovered by James Gregory.^[1]

By partial fractions and a substitution (Barrow's approach)

Although Gregory proved the conjecture in 1668 in his Exercitationes Geometricae,^[7] 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,^[8] gave the first "intelligible" proof, though even that was "couched in the geometric idiom of the day."^[3] Barrow's proof of the result was the earliest use of partial fractions in integration.^[3] Adapted to modern notation, Barrow's proof began as follows:

\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

Substituting $u = sin θ$ , $du = cos θ dθ$ , reduces the integral to

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

Therefore,

\int \sec \theta \,d\theta ={\frac {1}{2}}\ln {\frac {1+\sin \theta }{1-\sin \theta }}+C,

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

By the tangent half-angle substitution

Standard

Under the tangent half-angle substitution ${\textstyle t=\tan {\tfrac {1}{2}}\theta ,}$ ^[9]

{\begin{aligned}&\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 ={\frac {1}{\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{aligned}}

Therefore the integral of the secant function is

{\begin{aligned}\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{aligned}}

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,^[10] is as follows:

{\begin{aligned}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&={\frac {1}{2}}\sec ^{2}\left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)d\theta ={\frac {1}{2}}\left(1+x^{2}\right)d\theta \\[10pt]d\theta &={\frac {2}{1+x^{2}}}\,dx.\end{aligned}}

Substituting:

{\begin{aligned}\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{aligned}}

By two successive substitutions

The integral can also be solved by manipulating the integrand and substituting twice. Using the definition $sec θ = 1 / cos θ$ and the identity $cos 2 θ + sin 2 θ = 1$ , the integral can be rewritten as

\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 .

Substituting $u = sin θ$ , $du = cos θ dθ$ reduces the integral to

\int {\frac {1}{1-u^{2}}}\,du.

The reduced integral can be evaluated by substituting $u = tanh t$ , $du = sech 2 t dt$ , and then using the identity $1 - tanh 2 t = sech 2 t$ .

\int {\frac {\operatorname {sech} ^{2}t}{1-\tanh ^{2}t}}\,dt=\int {\frac {\operatorname {sech} ^{2}t}{\operatorname {sech} ^{2}t}}\,dt=\int dt.

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

{\begin{aligned}\int dt&=t+C\\&=\operatorname {artanh} u+C\\[4pt]&=\operatorname {artanh} (\sin \theta )+C,\end{aligned}}

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:

\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 ,

where $\pm$ stands for $\operatorname {sgn}(\cos \theta )$ because ${\sqrt {1+\tan ^{2}\theta }}=|\sec \theta \,|.$ Substituting $u = tan θ$ , $du = sec 2 θ dθ$ , reduces to a standard integral:

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

where $sgn$ is the sign function.

Likewise:

\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 .

Substituting $u = | sec θ |$ , $du = | sec θ | tan θ dθ$ , reduces to a standard integral:

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

Using complex exponential form

Under the substitution $z=e^{i\theta },$

{\begin{aligned}&\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 ={\frac {2}{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{aligned}}

So the integral can be solved as:

{\begin{aligned}\int \sec \theta \,d\theta &=\int {\frac {2}{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{aligned}}

Because the constant of integration can be anything, the additional constant term can be absorbed into it. Finally, if theta is real-valued, we can indicate this with absolute value brackets in order to get the equation into its most familiar form:

\int \sec \theta \,d\theta =\ln \left|\tan \theta +\sec \theta \right|+C

Gudermannian and Lambertian

The integral of the hyperbolic secant function defines the Gudermannian function:

\int _{0}^{\psi }\operatorname {sech} u\,du=\operatorname {gd} \psi .

The integral of the secant function defines the Lambertian function, which is the inverse of the Gudermannian function:

\int _{0}^{\varphi }\sec t\,dt=\operatorname {lam} \varphi =\operatorname {gd} ^{-1}\varphi .

These functions are encountered in the theory of map projections: the Mercator projection of a point on the sphere with longitude $λ$ and latitude $ϕ$ may be written^[11] as:

(x,y)={\bigl (}\lambda ,\operatorname {lam} \varphi {\bigr )}.

Notes

^ ^a ^b ^c Stewart, James (2012). "Section 7.2: Trigonometric Integrals". Calculus - Early Transcendentals. Cengage Learning. pp. 475–6. ISBN 978-0-538-49790-9.
^ For example this form is used in Karney, Charles F.F. (2011). "Transverse Mercator with an accuracy of a few nanometers". Journal of Geodesy. 85: 475–485.
^ ^a ^b ^c ^d ^e V. Frederick Rickey and Philip M. Tuchinsky, 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.
^ 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.
^ 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.
D. T. Whiteside, editor, The Mathematical Papers of Isaac Newton, Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.
^ Feldman, Joel. "Integration of sec x and sec³ x" (PDF). University of British Columbia Mathematics Department.
"Integral of Secant". MIT OpenCourseWare.
^ Gregory, James (1668). "Analogia Inter Lineam Meridianam Planispherii Nautici & Tangentes Artificiales Geometricè Demonstrata, &c." [Analogy Between the Meridian Line of the Nautical Planisphere & Artificial Tangents Geometrically Demonstrated, &c.]. Exercitationes Geometricae [Geometrical Exercises] (in Latin). Moses Pitt. pp. 14–24.
^ Barrow, Isaac (1674) [1670]. "Lectiones geometricae: XII, Appendicula I". Lectiones Opticae & Geometricae (in Latin). Typis Guilielmi Godbid. pp. 110–114. In English, "Lecture XII, Appendix I". The Geometrical Lectures of Isaac Barrow. Translated by Child, James Mark. Open Court. 1916. pp. 165–169.
^ Stewart, James (2012). "Section 7.4: Integration of Rational Functions by Partial Fractions". Calculus: Early Transcendentals (7th ed.). Belmont, CA, USA: Cengage Learning. pp. 493. ISBN 978-0-538-49790-9.
^ Hardy, Michael (2013). "Efficiency in Antidifferentiation of the Secant Function". American Mathematical Monthly. 120 (6): 580.
^ Lee, Laurence Patrick (1976). Conformal Projections Based on Elliptic Functions. Cartographica Monograph. Vol. 16. University of Toronto Press. ISBN 9780919870161.

References

Maseres, Francis, ed. (1791). Scriptores Logarithmici; Or, A Collection of Several Curious Tracts on the Nature and Construction of Logarithms. Vol. 2. J. Davis.
Strauss, Simon W. (1980). "The Integrals ${\textstyle \int \sec x\,dx}$ and ${\textstyle \int \csc x\,dx}$ Revisited". Journal of the Washington Academy of Sciences. 70 (4): 137–143. JSTOR 24537231.

[stewart-1] Stewart, James (2012). "Section 7.2: Trigonometric Integrals". Calculus - Early Transcendentals. Cengage Learning. pp. 475–6. ISBN 978-0-538-49790-9.

[2] For example this form is used in Karney, Charles F.F. (2011). "Transverse Mercator with an accuracy of a few nanometers". Journal of Geodesy. 85: 475–485.

[rickey-tuchinsky-3] V. Frederick Rickey and Philip M. Tuchinsky, 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.

[wright-4] 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.

[newton-5] 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.
D. T. Whiteside, editor, The Mathematical Papers of Isaac Newton, Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.

[handouts-6] Feldman, Joel. "Integration of sec x and sec³ x" (PDF). University of British Columbia Mathematics Department.
"Integral of Secant". MIT OpenCourseWare.

[gregory-7] Gregory, James (1668). "Analogia Inter Lineam Meridianam Planispherii Nautici & Tangentes Artificiales Geometricè Demonstrata, &c." [Analogy Between the Meridian Line of the Nautical Planisphere & Artificial Tangents Geometrically Demonstrated, &c.]. Exercitationes Geometricae [Geometrical Exercises] (in Latin). Moses Pitt. pp. 14–24.

[barrow-8] Barrow, Isaac (1674) [1670]. "Lectiones geometricae: XII, Appendicula I". Lectiones Opticae & Geometricae (in Latin). Typis Guilielmi Godbid. pp. 110–114. In English, "Lecture XII, Appendix I". The Geometrical Lectures of Isaac Barrow. Translated by Child, James Mark. Open Court. 1916. pp. 165–169.

[stewart2-9] Stewart, James (2012). "Section 7.4: Integration of Rational Functions by Partial Fractions". Calculus: Early Transcendentals (7th ed.). Belmont, CA, USA: Cengage Learning. pp. 493. ISBN 978-0-538-49790-9.

[hardy-10] Hardy, Michael (2013). "Efficiency in Antidifferentiation of the Secant Function". American Mathematical Monthly. 120 (6): 580.

[lee-11] Lee, Laurence Patrick (1976). Conformal Projections Based on Elliptic Functions. Cartographica Monograph. Vol. 16. University of Toronto Press. ISBN 9780919870161.

[1]

[2]

[3]

[4]

[5]

[6]

[7]

[8]

[9]

[10]

[11]

Calculus
Precalculus	Binomial theorem Concave function Continuous function Factorial Finite difference Free variables and bound variables Graph of a function Linear function Radian Rolle's theorem Secant Slope Tangent
Limits	Indeterminate form Limit of a function One-sided limit Limit of a sequence Order of approximation (ε, δ)-definition of limit
Differential calculus	Derivative Second derivative Partial derivative Differential Differential operator Mean value theorem Notation Leibniz's notation Newton's notation Rules of differentiation linearity Power Sum Chain L'Hôpital's Product General Leibniz's rule Quotient Other techniques Implicit differentiation Inverse functions and differentiation Logarithmic derivative Related rates Stationary points First derivative test Second derivative test Extreme value theorem Maximum and minimum Further applications Newton's method Taylor's theorem Differential equation Ordinary differential equation Partial differential equation Stochastic differential equation
Integral calculus	Antiderivative Arc length Riemann integral Basic properties Constant of integration Fundamental theorem of calculus Differentiating under the integral sign Integration by parts Integration by substitution trigonometric Euler Tangent half-angle substitution Partial fractions in integration Quadratic integral Trapezoidal rule Volumes Washer method Shell method Integral equation Integro-differential equation
Vector calculus	Derivatives Curl Directional derivative Divergence Gradient Laplacian Basic theorems Line integrals Green's Stokes' Gauss'
Multivariable calculus	Divergence theorem Geometric Hessian matrix Jacobian matrix and determinant Lagrange multiplier Line integral Matrix Multiple integral Partial derivative Surface integral Volume integral Advanced topics Differential forms Exterior derivative Generalized Stokes' theorem Tensor calculus
Sequences and series	Arithmetico-geometric sequence Types of series Alternating Binomial Fourier Geometric Harmonic Infinite Power Maclaurin Taylor Telescoping Tests of convergence Abel's Alternating series Cauchy condensation Direct comparison Dirichlet's Integral Limit comparison Ratio Root Term
Special functions and numbers	Bernoulli numbers e (mathematical constant) Exponential function Natural logarithm Stirling's approximation
History of calculus	Adequality Brook Taylor Colin Maclaurin Generality of algebra Gottfried Wilhelm Leibniz Infinitesimal Infinitesimal calculus Isaac Newton Fluxion Law of Continuity Leonhard Euler Method of Fluxions The Method of Mechanical Theorems
Lists	Differentiation rules List of integrals of exponential functions List of integrals of hyperbolic functions List of integrals of inverse hyperbolic functions List of integrals of inverse trigonometric functions List of integrals of irrational functions List of integrals of logarithmic functions List of integrals of rational functions List of integrals of trigonometric functions Secant Secant cubed List of limits Lists of integrals
Miscellaneous topics	Complex calculus Contour integral Differential geometry Manifold Curvature of curves of surfaces Tensor Euler–Maclaurin formula Gabriel's horn Integration Bee Proof that 22/7 exceeds π Regiomontanus' angle maximization problem Steinmetz solid