Integral of the secant function

Three common expressions for the integral of the secant,

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta &={\dfrac {1}{2}}\ln {\dfrac {1+\sin \theta }{1-\sin \theta }}+C\\[5mu]&=\ln {{\bigl |}\sec \theta +\tan \theta \,{\bigr |}}+C\\[5mu]&=\ln {\left|\,{\tan }{\biggl (}{\frac {\theta }{2}}+{\frac {\pi }{4}}{\biggr )}\right|}+C,\end{aligned}}}$

are equivalent because

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

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

First,

${\displaystyle {\begin{aligned}{\sqrt {\dfrac {1+\sin \theta }{1-\sin \theta }}}&={\sqrt {\frac {1+{\dfrac {2t}{1+t^{2}}}}{1-{\dfrac {2t}{1+t^{2}}}}}}={\sqrt {\frac {1+t^{2}+2t}{1+t^{2}-2t}}}={\sqrt {\frac {(1+t)^{2}}{(1-t)^{2}}}}\\[5mu]&=\left|{\frac {1+t}{1-t}}\right|.\end{aligned}}}$

Second,

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "http://localhost:6011/en.wikipedia.org/v1/":): {\displaystyle \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 ${\displaystyle \tan(\phi +\psi )=(\tan \phi +\tan \psi ){\big /}(1-\tan \phi \,\tan \psi ),}$

${\displaystyle {\begin{aligned}\left|\,{\tan }{\biggl (}{\frac {\theta }{2}}+{\frac {\pi }{4}}{\biggr )}\right|&=\left|{\frac {\tan {\tfrac {1}{2}}\theta +\tan {\tfrac {1}{4}}\pi }{1-\tan {\tfrac {1}{2}}\theta \,\tan {\tfrac {1}{4}}\pi }}\right|=\left|{\frac {t+1}{1-t\cdot 1}}\right|\\[5mu]&=\left|{\frac {1+t}{1-t}}\right|.\end{aligned}}}$

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 ${\displaystyle \varphi }$ lies between ${\textstyle -{\tfrac {1}{2}}\pi }$ and ${\textstyle {\tfrac {1}{2}}\pi }$,

${\displaystyle y=\ln \,{\tan }{\biggl (}{\frac {\varphi }{2}}+{\frac {\pi }{4}}{\biggr )}.}$

Let

${\displaystyle {\begin{aligned}\psi &=\ln(\sec \theta +\tan \theta ),\\[4pt]e^{\psi }&=\sec \theta +\tan \theta ,\\[4pt]\sinh \psi &={\frac {e^{\psi }-e^{-\psi }}{2}}=\tan \theta ,\\[4pt]\cosh \psi &={\sqrt {1+\sinh ^{2}\psi }}=|\sec \theta \,|,\\[4pt]\tanh \psi &=\sin \theta .\end{aligned}}}$

Therefore,

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta &=\operatorname {artanh} \left(\sin \theta \right)+C\\[-2mu]&=\operatorname {sgn}(\cos \theta )\operatorname {arsinh} \left(\tan \theta \right)+C\\[7mu]&=\operatorname {sgn}(\sin \theta )\operatorname {arcosh} {\left|\sec \theta \right|}+C.\end{aligned}}}$

A standard method of evaluating the secant integral presented in various references involves multiplying the numerator and denominator by sec θ + tan θ and then using the substitution 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.[6]

Starting with

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

adding them gives

${\displaystyle {\begin{aligned}{\frac {d}{d\theta }}(\sec \theta +\tan \theta )&=\sec \theta \tan \theta +\sec ^{2}\theta \\&=\sec \theta (\tan \theta +\sec \theta ).\end{aligned}}}$

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

${\displaystyle {\begin{aligned}u&=\sec \theta +\tan \theta \\du&=\left(\sec \theta \tan \theta +\sec ^{2}\theta \right)\,d\theta .\end{aligned}}}$

The integral is evaluated as follows:

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta &=\int {\frac {\sec \theta (\sec \theta +\tan \theta )}{\sec \theta +\tan \theta }}\,d\theta \\[6pt]&=\int {\frac {\sec ^{2}\theta +\sec \theta \tan \theta }{\sec \theta +\tan \theta }}\,d\theta &u&=\sec \theta +\tan \theta \\[6pt]&=\int {\frac {1}{u}}\,du&du&=\left(\sec \theta \tan \theta +\sec ^{2}\theta \right)\,d\theta \\[6pt]&=\ln |u|+C\\[4pt]&=\ln |\sec \theta +\tan \theta |+C,\end{aligned}}}$

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

Although Gregory proved the conjecture in 1668 in his Exercitationes Geometricae,[7] 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,[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 of partial fractions in integration.[3] Adapted to modern notation, Barrow's proof began as follows:

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

Substituting u = sin θ, du = cos θ dθ, reduces the integral to

${\displaystyle {\begin{aligned}\int {\frac {1}{1-u^{2}}}\,du&=\int {\frac {1}{(1+u)(1-u)}}\,du\\[6pt]&=\int {\frac {1}{2}}\!\left({\frac {1}{1+u}}+{\frac {1}{1-u}}\right)du&&{\text{partial fraction decomposition}}\\[6pt]&={\frac {1}{2}}{\bigl (}\ln \left|1+u\right|-\ln \left|1-u\right|{\bigr )}+C\\[6pt]&={\frac {1}{2}}\ln \left|{\frac {1+u}{1-u}}\right|+C\end{aligned}}}$

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 the tangent half-angle substitution ${\textstyle t=\tan {\tfrac {1}{2}}\theta ,}$[9]

${\displaystyle {\begin{aligned}&\sin \theta ={\frac {2t}{1+t^{2}}},\quad \cos \theta ={\frac {1-t^{2}}{1+t^{2}}},\quad d\theta ={\frac {2}{1+t^{2}}}\,dt,\\[10mu]&\tan \theta ={\frac {\sin \theta }{\cos \theta }}={\frac {2t}{1-t^{2}}},\quad \sec \theta ={\frac {1}{\cos \theta }}={\frac {1+t^{2}}{1-t^{2}}},\\[10mu]&\sec \theta +\tan \theta ={\frac {1+2t+t^{2}}{1-t^{2}}}={\frac {1+t}{1-t}}.\end{aligned}}}$

Therefore the integral of the secant function is

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta &=\int \left({\frac {1+t^{2}}{1-t^{2}}}\right)\!\left({\frac {2}{1+t^{2}}}\right)dt&&t=\tan {\frac {\theta }{2}}\\[6pt]&=\int {\frac {2}{(1-t)(1+t)}}\,dt\\[6pt]&=\int \left({\frac {1}{1+t}}+{\frac {1}{1-t}}\right)dt&&{\text{partial fraction decomposition}}\\[6pt]&=\ln |1+t|-\ln |1-t|+C\\[6pt]&=\ln \left|{\frac {1+t}{1-t}}\right|+C\\[6pt]&=\ln |\sec \theta +\tan \theta |+C,\end{aligned}}}$

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:

${\displaystyle {\begin{aligned}x&=\tan \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\\[10pt]{\frac {2x}{1+x^{2}}}&={\frac {2\tan \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)}{\sec ^{2}\left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)}}=2\sin \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\cos \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\\[6pt]&=\sin \left({\frac {\pi }{2}}+\theta \right)=\cos \theta &&{\text{by the double-angle formula}}\\[10pt]dx&={\frac {1}{2}}\sec ^{2}\left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)d\theta ={\frac {1}{2}}\left(1+x^{2}\right)d\theta \\[10pt]d\theta &={\frac {2}{1+x^{2}}}\,dx.\end{aligned}}}$

Substituting:

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta =\int {\frac {1}{\cos \theta }}\,d\theta &=\int {\frac {1+x^{2}}{2x}}\cdot {\frac {2}{1+x^{2}}}\,dx\\[6pt]&=\int {\frac {1}{x}}\,dx\\[6pt]&=\ln |x|+C\\[6pt]&=\ln \left|\tan \left({\frac {\pi }{4}}+{\frac {\theta }{2}}\right)\right|+C.\end{aligned}}}$

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

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

Substituting u = sin θ, du = cos θ dθ reduces the integral to

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

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

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

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

${\displaystyle {\begin{aligned}\int dt&=t+C\\&=\operatorname {artanh} u+C\\[4pt]&=\operatorname {artanh} (\sin \theta )+C,\end{aligned}}}$

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:

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

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

${\displaystyle {\begin{aligned}\int {\frac {1}{\pm {\sqrt {1+u^{2}}}}}\,du&=\pm \operatorname {arsinh} u+C\\&=\operatorname {sgn}(\cos \theta )\operatorname {arsinh} \left(\tan \theta \right)+C,\end{aligned}}}$

where sgn is the sign function.

Likewise:

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

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

${\displaystyle {\begin{aligned}\int {\frac {1}{\pm {\sqrt {u^{2}-1}}}}\,du&=\pm \operatorname {arcosh} u+C\\&=\operatorname {sgn}(\sin \theta )\operatorname {arcosh} \left|\sec \theta \right|+C.\end{aligned}}}$

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

${\displaystyle {\begin{aligned}&\theta =-i\ln z,\quad d\theta ={\frac {-i}{z}}dz,\quad \cos \theta ={\frac {z+z^{-1}}{2}},\quad \sin \theta ={\frac {z-z^{-1}}{2i}},\quad \\[5mu]&\sec \theta ={\frac {2}{z+z^{-1}}},\quad \tan \theta =-i{\frac {z-z^{-1}}{z+z^{-1}}},\quad \\[5mu]&\sec \theta +\tan \theta =-i{\frac {2i+z-z^{-1}}{z+z^{-1}}}=-i{\frac {(z+i)(1+iz^{-1})}{(z-i)(1+iz^{-1})}}=-i{\frac {z+i}{z-i}}\end{aligned}}}$

So the integral can be solved as:

${\displaystyle {\begin{aligned}\int \sec \theta \,d\theta &=\int {\frac {2}{z+z^{-1}}}\,{\frac {-i}{z}}dz&&z=e^{i\theta }\\[5mu]&=\int {\frac {-2i}{z^{2}+1}}dz\\&=\int {\frac {1}{z+i}}-{\frac {1}{z-i}}\,dz&&{\text{partial fraction decomposition}}\\[5mu]&=\ln(z+i)-\ln(z-i)+C\\[5mu]&=\ln {\frac {z+i}{z-i}}+C\\[5mu]&=\ln {\bigl (}i(\sec \theta +\tan \theta ){\bigr )}+C\\[5mu]&=\ln(\sec \theta +\tan \theta )+\ln i+C\end{aligned}}}$

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:

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