integral of the secant function

Three common expressions for the integral of the secant,

:$\backslash begin\{align\}$

\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}

are equivalent because

:$$

\sqrt{\dfrac{1+\sin\theta}{1-\sin\theta}}

= \bigl|\sec\theta + \tan\theta\,\bigr|

= \left|\,{\tan}\biggl(\frac\theta2 + \frac\pi4\biggr) \right|.

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

First,

:$\backslash begin\{align\}$

\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}

Second,

:$\backslash begin\{align\}$

\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}

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

:$\backslash begin\{align\}$

\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}

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 $\backslash varphi$ lies between $-\backslash tfrac12\backslash pi$ and $\backslash tfrac12\backslash pi$,

:$y\; =\; \backslash ln\backslash ,\{\backslash tan\}\backslash biggl(\backslash frac\backslash varphi2\; +\; \backslash frac\backslash pi4\backslash biggr).$

Let

:$$

\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}

Therefore,

:$\backslash begin\{align\}$

\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}

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

:$\backslash int\_0^\backslash varphi\; \backslash sec\backslash theta\backslash ,d\backslash theta\; =\; \backslash ln\backslash tan\backslash left(\backslash frac\{\backslash varphi\}\{2\}\; +\; \backslash frac\{\backslash pi\}\{4\}\backslash right).$

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

A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator by {{math|sec θ + tan θ}} and then using the substitution {{math|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.

Starting with

:$\backslash frac\; d\{d\backslash theta\}\backslash sec\; \backslash theta\; =\; \backslash sec\backslash theta\; \backslash tan\backslash theta\; \backslash quad\; \backslash text\{and\}\; \backslash quad\; \backslash frac\; d\{d\backslash theta\}\backslash tan\; \backslash theta\; =\; \backslash sec^2\; \backslash theta,$

adding them gives

:$\backslash begin\{align\}$

\frac d{d\theta}(\sec\theta + \tan\theta) &= \sec\theta \tan\theta + \sec^2\theta \\

&= \sec\theta (\tan\theta + \sec \theta).

\end{align}

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

:$\backslash begin\{align\}$

u &= \sec \theta + \tan \theta \\

du &= \left(\sec \theta \tan \theta + \sec^2 \theta\right)\,d\theta.

\end{align}

The integral is evaluated as follows:

: $\backslash begin\{align\}$

\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}

as claimed. This was the formula discovered by James Gregory.

Although Gregory proved the conjecture in 1668 in his Exercitationes Geometricae, 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, gave the first "intelligible" proof, though even that was "couched in the geometric idiom of the day." Barrow's proof of the result was the earliest use of partial fractions in integration. 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 {{math|u {{=}} sin θ}}, {{math|du {{=}} cos θ dθ}}, reduces the integral to

:$\backslash begin\{align\}$

\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}

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 $1\; +\; \backslash sin\; \backslash theta$ and $1\; -\; \backslash sin\; \backslash theta$ are always non-negative for real values of $\backslash theta.$

Under the tangent half-angle substitution $t\; =\; \backslash tan\backslash tfrac12\backslash theta,$

:$\backslash begin\{align\}$

&\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}

Therefore the integral of the secant function is

:$\backslash begin\{align\}$

\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}

as before.

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, is as follows:

: $\backslash begin\{align\}$

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}

Substituting:

: $\backslash begin\{align\}$

\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}

The integral can also be solved by manipulating the integrand and substituting twice. Using the definition {{math|sec θ {{=}} {{sfrac|1|cos θ}}}} and the identity {{math|cos² θ + sin² θ {{=}} 1}}, the integral can be rewritten as

: $\backslash int\; \backslash sec\; \backslash theta\; \backslash ,\; d\backslash theta=\backslash int\; \backslash frac\{1\}\{\backslash cos\; \backslash theta\}\backslash ,d\backslash theta\; =\; \backslash int\; \backslash frac\{\backslash cos\; \backslash theta\}\{\backslash cos^2\; \backslash theta\}\backslash ,d\backslash theta=\backslash int\; \backslash frac\{\backslash cos\; \backslash theta\}\{1-\backslash sin^2\; \backslash theta\}\backslash ,d\backslash theta.$

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

: $\backslash int\; \backslash frac\{1\}\{1-u^2\}\; \backslash ,\; du.$

The reduced integral can be evaluated by substituting {{math|u {{=}} tanh t}}, {{math|du {{=}} sech² t dt}}, and then using the identity {{math|1 − tanh² t {{=}} sech² t}}.

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

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

:$\backslash begin\{align\}$

\int dt &= t+C \\

&=\operatorname{artanh} u+C \\[4pt]

&=\operatorname{artanh}(\sin \theta)+C,

\end{align}

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.

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

:$\backslash int\; \backslash sec\; \backslash theta\; \backslash ,d\backslash theta=\backslash int\; \backslash frac\{\backslash sec^2\; \backslash theta\}\{\backslash sec\; \backslash theta\}\; \backslash ,d\backslash theta=\backslash int\; \backslash frac\{\backslash sec^2\; \backslash theta\}\{\backslash pm\backslash sqrt\{1+\backslash tan^2\; \backslash theta\}\}\; \backslash ,d\backslash theta,$

where $\backslash pm$ stands for $\backslash sgn(\backslash cos\; \backslash theta)$ because $\backslash sqrt\{1+\backslash tan^2\; \backslash theta\}\; =\; |\backslash sec\backslash theta\backslash ,|.$ Substituting {{math|u {{=}} tan θ}}, {{math|du {{=}} sec² θ dθ}}, reduces to a standard integral:

:$\backslash begin\{align\}$

\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}

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

Likewise:

:$\backslash int\; \backslash sec\; \backslash theta\; \backslash ,d\backslash theta\; =\backslash int\; \backslash frac\{\backslash sec\; \backslash theta\; \backslash tan\; \backslash theta\}\{\backslash tan\; \backslash theta\}\; \backslash ,d\backslash theta=\backslash int\; \backslash frac\{\backslash sec\; \backslash theta\; \backslash tan\; \backslash theta\}\{\backslash pm\backslash sqrt\{\backslash sec^2\; \backslash theta-1\}\}\; \backslash ,d\backslash theta.$

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

:$\backslash begin\{align\}$

\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}

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

:$\backslash begin\{align\}$

&\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}

So the integral can be solved as:

:$\backslash begin\{align\}$

\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}

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

File:Gudermannian function.png 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 {{sfrac|1|2}}ϕ {{=}} tanh {{sfrac|1|2}}ψ.}} The blue point has coordinates {{math|(cosh ψ, sinh ψ)}}. The red point has coordinates {{math|(cos ϕ, sin ϕ).}} The purple point has coordinates {{math|(0, s).}} ]]

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

:$\backslash int\_0^\backslash psi\; \backslash operatorname\{sech\}\; u\; \backslash ,\; du$

=\operatorname{gd}\psi.

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

:$\backslash int\_0^\backslash varphi\; \backslash sec\; t\; \backslash ,\; dt$

=\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 {{mvar|λ}} and latitude {{mvar|ϕ}} may be written as:

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

{{Portal|Mathematics}}

• Integral of secant cubed

{{clear}}

{{reflist |30em |refs=

{{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/

}}

{{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

}}

{{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

}}

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.

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.

{{cite web

| last = Feldman

| first = Joel

| title = Integration of sec x and sec³ x

| url = https://www.math.ubc.ca/~feldman/m121/secx.pdf

| website = University of British Columbia Mathematics Department

}}

{{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

}}

{{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

}}

{{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

}}

{{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].

}}

• {{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 $\backslash int\backslash sec\; x\backslash ,dx$ and $\backslash int\backslash csc\; x\backslash ,dx$ Revisited |journal=Journal of the Washington Academy of Sciences |volume=70 |number=4 |pages=137–143 |jstor=24537231 }}

{{Calculus topics}}

Category:Integral calculus