Incalculus, theintegral of the secant function can be evaluated using a variety of methods and there are multiple ways of expressing theantiderivative, all of which can be shown to be equivalent viatrigonometric identities,

${\displaystyle \int \sec \theta d\theta =\{\begin{array}{l}{\displaystyle \frac{1}{2}}\ln {\displaystyle \frac{1+\sin \theta }{1-\sin \theta }}+C\\ \ln |\sec \theta +\tan \theta |+C\\ \ln \left|\tan ({\displaystyle \frac{\theta }{2}}+{\displaystyle \frac{\pi }{4}})\right|+C\end{array}}$

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

The definite integral of the secant function starting from${\displaystyle 0}$ is the inverseGudermannian function,${\mathrm{gd}}^{-1}.$ For numerical applications, all of the above expressions result inloss of significance for some arguments. An alternative expression in terms of theinverse hyperbolic sinearsinh is numerically well behaved for real arguments:^[2]

${\displaystyle {\mathrm{gd}}^{-1}\varphi ={\int }_0^{\varphi }\sec \theta d\theta =\mathrm{arsinh}(\tan \varphi ).}$

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 theMercator projection, used formarine navigation with constant compass bearing.

Three common expressions for the integral of the secant,

are equivalent because

${\displaystyle \sqrt{{\displaystyle \frac{1+\sin \theta }{1-\sin \theta }}}=|\sec \theta +\tan \theta |=\left|\tan (\frac{\theta }{2}+\frac{\pi }{4})\right|.}$

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

First,

Second,

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

So all three expressions describe the same quantity.

The conventional solution for theMercator projection ordinate may be written without theabsolute value signs since the latitude${\displaystyle \phi }$ lies between$-\frac{1}{2}\pi$ and$\frac{1}{2}\pi$,

${\displaystyle y=\ln \tan (\frac{\phi }{2}+\frac{\pi }{4}).}$

Let

Therefore,

Theintegral of thesecant function was one of the "outstanding open problems of the mid-seventeenth century", solved in 1668 byJames Gregory.^[3] He applied his result to a problem concerning nautical tables.^[1] In 1599,Edward Wright evaluated the integral bynumerical methods – what today we would callRiemann sums.^[4] He wanted the solution for the purposes ofcartography – specifically for constructing an accurateMercator 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 consequentlyconjectured that^[3]

${\displaystyle {\int }_0^{\phi }\sec \theta d\theta =\ln \tan \left(\frac{\phi }{2}+\frac{\pi }{4}\right).}$

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

A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator bysec θ + tan θ and then using the substitutionu = sec θ + tan θ. This substitution can be obtained from thederivatives of secant and tangent added together, which have secant as a common factor.^[6]

Starting with

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

adding them gives

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

The integral is evaluated as follows:

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

Although Gregoryproved the conjecture in 1668 in hisExercitationes Geometricae,^[7] the proof was presented in a form that renders it nearly impossible for modern readers to comprehend;Isaac Barrow, in hisLectiones 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 ofpartial fractions in integration.^[3] Adapted to modern notation, Barrow's proof began as follows:

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

Substitutingu = sin θ,du = cos θdθ, reduces the integral to

Therefore,

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

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

Under thetangent half-angle substitution$t=\tan \frac{1}{2}\theta ,$^[9]

Therefore the integral of the secant function is

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,^[10] is as follows:

Substituting:

The integral can also be solved by manipulating theintegrand and substituting twice. Using the definitionsec θ =⁠1/cos θ⁠ and the identitycos² θ + sin² θ = 1, the integral can be rewritten as

${\displaystyle \int \sec \theta d\theta =\int \frac{1}{\cos \theta }d\theta =\int \frac{\cos \theta }{{\cos }^2\theta }d\theta =\int \frac{\cos \theta }{1-{\sin }^2\theta }d\theta .}$

Substitutingu = sin θ,du = cos θdθ reduces the integral to

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

The reduced integral can be evaluated by substitutingu = tanh t,du = sech² tdt, and then using the identity1 − tanh² t = sech² t.

${\displaystyle \int \frac{{\mathrm{sech}}^{2}t}{1-{\tanh }^{2}t}dt=\int \frac{{\mathrm{sech}}^{2}t}{{\mathrm{sech}}^{2}t}dt=\int dt.}$

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

which is one of the hyperbolic forms of the integral.

A similar strategy can be used to integrate thecosecant,hyperbolic secant, andhyperbolic cosecant functions.

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

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

where${\displaystyle \pm }$ stands for${\displaystyle \mathrm{sgn}(\cos \theta )}$ because${\displaystyle \sqrt{1+{\tan }^2\theta }=|\sec \theta |.}$ Substitutingu = tan θ,du = sec² θdθ, reduces to a standard integral:

wheresgn is thesign function.

Likewise:

${\displaystyle \int \sec \theta d\theta =\int \frac{\sec \theta \tan \theta }{\tan \theta }d\theta =\int \frac{\sec \theta \tan \theta }{\pm \sqrt{{\sec }^2\theta -1}}d\theta .}$

Substitutingu = |sec θ|,du = |sec θ| tan θdθ, reduces to a standard integral:

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

So the integral can be solved as:

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

${\displaystyle \int \sec \theta d\theta =\ln \left|\tan \theta +\sec \theta \right|+C}$

The integral of the hyperbolic secant function defines theGudermannian function:

${\displaystyle {\int }_0^{\psi }\mathrm{sech}udu=\mathrm{gd}\psi .}$

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

${\displaystyle {\int }_0^{\phi }\sec tdt=\mathrm{lam}\phi ={\mathrm{gd}}^{-1}\phi .}$

These functions are encountered in the theory of map projections: theMercator projection of a point on thesphere with longitudeλ and latitudeϕ may be written^[11] as:

${\displaystyle (x,y)=(\lambda ,\mathrm{lam}\phi ).}$

• Integral of secant cubed

• 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$\int \sec xdx$ and$\int \csc xdx$ Revisited".Journal of the Washington Academy of Sciences.70 (4):137–143.JSTOR 24537231.

• Integral calculus

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