# Integral of the secant function

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The integral of the secant function of trigonometry was the subject of one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 by James Gregory. 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

$\int _{0}^{\theta }\sec \theta \,d\theta =\ln \left|\tan \left({\frac {\theta }{2}}+{\frac {\pi }{4}}\right)\right|.$ That conjecture became widely known, and in 1665, Isaac Newton was aware of it.

## Evaluations

### Barrow's approach

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 Geometrical Lectures 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:

$\int \sec \theta \,d\theta =\int {\frac {d\theta }{\cos \theta }}=\int {\frac {\cos \theta \,d\theta }{\cos ^{2}\theta }}=\int {\frac {\cos \theta \,d\theta }{1-\sin ^{2}\theta }}$ Substituting $u$ for $\sin \theta$ reduces the integral to

{\begin{aligned}\int {\frac {du}{1-u^{2}}}&=\int {\frac {du}{(1-u)(1+u)}}={\dfrac {1}{2}}\int \left({\frac {1}{1+u}}+{\frac {1}{1-u}}\right)\,du\\[10pt]&={\frac {1}{2}}\left(\ln \left|1+u\right|-\ln \left|1-u\right|\right)+C={\frac {1}{2}}\ln \left|{\frac {1+u}{1-u}}\right|+C\end{aligned}} Therefore,

$\int \sec \theta \,d\theta =\left\{{\begin{array}{l}{\dfrac {1}{2}}\ln \left|{\dfrac {1+\sin \theta }{1-\sin \theta }}\right|+C\\[15pt]\ln \left|\sec \theta +\tan \theta \right|+C\\[15pt]\ln \left|\tan \left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)\right|+C\end{array}}\right\}{\text{ (equivalent forms)}}$ The second of these follows by first multiplying top and bottom of the interior fraction by $(1+\sin \theta )$ . This gives $\cos ^{2}\theta$ in the denominator and the result follows by moving the factor of 1/2 into the logarithm as a square root.

The third form follows by replacing $\sin \theta$ by $-\cos(\theta +\pi /2)$ and expanding using the identities for $\cos 2x$ . It may also be obtained directly by means of the following substitutions:

{\begin{aligned}\sec \theta ={\frac {1}{\sin \left(\theta +{\dfrac {\pi }{2}}\right)}}={\frac {1}{2\sin \left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)\cos \left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)}}={\frac {\sec ^{2}\left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)}{2\tan \left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)}}.\end{aligned}} The conventional solution for the Mercator projection ordinate may be written without the modulus signs since the latitude (φ) lies between −π/2 and π/2:

$y=\ln \tan \!\left({\frac {\varphi }{2}}+{\frac {\pi }{4}}\right).$ ### By the Euler substitution

The integral can also be derived by using the tangent half-angle substitution, also known as the Euler or Weierstrass substitution. 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:

{\begin{aligned}&x=\tan \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\\[10pt]&{\frac {2x}{1+x^{2}}}=2\sin \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\cos \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)=\sin \left({\frac {\pi }{2}}+\theta \right)=\cos \theta \\[10pt]&dx={\frac {1}{2}}\sec ^{2}\left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)d\theta ={\frac {1}{2}}(1+x^{2})d\theta \\[10pt]&{\frac {2\,dx}{1+x^{2}}}=d\theta \\[10pt]\int \sec \theta \,d\theta &=\int {\frac {dx}{x}}=\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\right|+C.\end{aligned}} ## Hyperbolic forms

Let

{\begin{aligned}\psi &=\ln(\sec \theta +\tan \theta ),\\e^{\psi }&=\sec \theta +\tan \theta ,\\\sinh \psi &={\frac {1}{2}}(e^{\psi }-e^{-\psi })=\tan \theta ,\\\cosh \psi &={\sqrt {1+\sinh ^{2}\psi }}=\sec \theta ,\\\tanh \psi &=\sin \theta .\end{aligned}} Therefore,

{\begin{aligned}\int \sec \theta \,d\theta &=\psi =\tanh ^{-1}\!\left(\sin \theta \right)=\sinh ^{-1}\!\left(\tan \theta \right)=\cosh ^{-1}\!\left(\sec \theta \right).\end{aligned}} ## Gudermannian and lambertian

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

{\begin{aligned}\int \sec \theta \,d\theta &=\operatorname {gd} ^{-1}(\theta )=\operatorname {lam} (\theta ).\end{aligned}} The lambertian function (lam) is a notation for the inverse of the gudermannian which is encountered in the theory of map projections. In particular the Mercator projection may be written as

$y=\operatorname {lam} (\varphi ).$ ## Notes and references

1. 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.
2. ^ 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.
3. ^ 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.
4. ^ D. T. Whiteside, editor, The Mathematical Papers of Isaac Newton, Cambridge University Press, 1967, volume 1, pages 466–467 and 473–475.
5. ^ Dresden, Arnold (1918). "Review: The Geometrical Lectures of Isaac Barrow, translated, with notes and proofs, by James Mark Child" (PDF). Bull. Amer. Math. Soc. 24 (9): 454–456. doi:10.1090/s0002-9904-1918-03122-4.
6. ^ Michael Hardy, "Efficiency in Antidifferentiation of the Secant Function", American Mathematical Monthly, June–July 2013, page 580.
7. ^ Lee, L.P. (1976). Conformal Projections Based on Elliptic Functions. Supplement No. 1 to Canadian Cartographer, Vol 13. (Designated as Monograph 16)