Greenwood frequency

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

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

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

The problem was solved by Isaac Barrow. His proof of the result was the earliest use of partial fractions in integration. Adapted to modern notation, Barrow's proof began as follows:

$\int \sec \theta \,d\theta =\int {\frac {d\theta }{\cos \theta }}=\int {\frac {\cos \theta \,d\theta }{\cos ^{2}\theta }}=\int {\frac {\cos \theta \,d\theta }{1-\sin ^{2}\theta }}=\int {\frac {du}{1-u^{2}}}$

This reduces it to the problem of antidifferentiating a rational function by using partial fractions. The proof goes on from there:

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

Finally, we convert it back to a function of θ:

$=\left\{{\begin{array}{l}{\dfrac {1}{2}}\ln \left|{\dfrac {1+\sin \theta }{1-\sin \theta }}\right|+C\\[15pt]\ln \left|\sec \theta +\tan \theta \right|+C\\[15pt]\ln \left|\tan \left({\dfrac {\theta }{2}}+{\dfrac {\pi }{4}}\right)\right|+C\end{array}}\right\}{\text{ (equivalent forms)}}$

The third form may 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:

$y=\ln \tan \!\left({\dfrac {\phi }{2}}+{\dfrac {\pi }{4}}\right).$

The problem can also be done by using the tangent half-angle substitution, but the details become somewhat more complicated than in the argument above.

Hyperbolic forms

Let

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

Therefore

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

Gudermannian and lambertian

${\begin{aligned}\int \sec \theta \,d\theta &={\mbox{gd}}^{-1}(\theta )={\mbox{lam}}(\theta ).\end{aligned}}$

gd is the Gudermannian function. The lambertian form (lam) is encountered in the theory of map projections.

Notes and references

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See also

• Integral of secant cubed
• Gudermannian function

External links

• Rickey and Tuchinsky's paper on the history of this integral