# 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

## Contents

0 θ sec ζ d ζ = ln | tan ( θ 2 + π 4 ) | .

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:

sec θ d θ = d θ cos θ = cos θ d θ cos 2 θ = cos θ d θ 1 sin 2 θ

Substituting u for sin θ reduces the integral to

d u 1 u 2 = d u ( 1 u ) ( 1 + u ) = 1 2 ( 1 1 + u + 1 1 u ) d u = 1 2 ln | 1 + u | 1 2 ln | 1 u | + C = 1 2 ln | 1 + u 1 u | + C

Therefore,

sec θ d θ = { 1 2 ln | 1 + sin θ 1 sin θ | + C ln | sec θ + tan θ | + C ln | tan ( θ 2 + π 4 ) | + C }  (equivalent forms)

The second of these follows by first multiplying top and bottom of the interior fraction by ( 1 + sin θ ) . This gives cos 2 θ 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 θ by cos ( θ + π / 2 ) and expanding using the identities for cos 2 x . It may also be obtained directly by means of the following substitutions:

sec θ = 1 sin ( θ + π 2 ) = 1 2 sin ( θ 2 + π 4 ) cos ( θ 2 + π 4 ) = sec 2 ( θ 2 + π 4 ) 2 tan ( θ 2 + π 4 ) .

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 ( φ 2 + π 4 ) .

The integral can also be derived by using the tangent half-angle 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:

x = tan ( π 4 + θ 2 ) 2 x x 2 + 1 = cos θ 2 d x x 2 + 1 = d θ sec θ d θ = d x x = log | tan ( π 4 + θ 2 ) | + constant .

## Hyperbolic forms

Let

ψ = ln ( sec θ + tan θ ) , e ψ = sec θ + tan θ , sinh ψ = 1 2 ( e ψ e ψ ) = tan θ , cosh ψ = 1 + sinh 2 ψ = sec θ , tanh ψ = sin θ .

Therefore,

sec θ d θ = ψ = tanh 1 ( sin θ ) = sinh 1 ( tan θ ) = cosh 1 ( sec θ ) .

## Gudermannian and lambertian

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

sec θ d θ = gd 1 ( θ ) = lam ( θ ) .

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 = lam ( φ ) .

## References

Integral of the secant function Wikipedia

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