Integral of secant cubed

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The integral of secant cubed is a frequent and challenging^[1] indefinite integral of elementary calculus:

${\textstyle {\begin{aligned}\int \sec ^{3}x,円dx&={\tfrac {1}{2}}{\frac {d}{dx}}\sec x+{\tfrac {1}{2}}\int \sec x,円dx+C\\[6mu]&={\tfrac {1}{2}}\sec x\tan x+{\tfrac {1}{2}}\int \sec x,円dx+C\\[6mu]&={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C\\[6mu]&={\tfrac {1}{2}}(\sec x\tan x+\operatorname {gd} ^{-1}x)+C,\qquad |x|<{\tfrac {1}{2}}\pi \end{aligned}}}${\textstyle {\begin{aligned}\int \sec ^{3}x,円dx&={\tfrac {1}{2}}{\frac {d}{dx}}\sec x+{\tfrac {1}{2}}\int \sec x,円dx+C\\[6mu]&={\tfrac {1}{2}}\sec x\tan x+{\tfrac {1}{2}}\int \sec x,円dx+C\\[6mu]&={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C\\[6mu]&={\tfrac {1}{2}}(\sec x\tan x+\operatorname {gd} ^{-1}x)+C,\qquad |x|<{\tfrac {1}{2}}\pi \end{aligned}}}

where ${\textstyle \operatorname {gd} ^{-1}}${\textstyle \operatorname {gd} ^{-1}} is the inverse Gudermannian function, the integral of the secant function.

There are a number of reasons why this particular antiderivative is worthy of special attention:

• The technique used for reducing integrals of higher odd powers of secant to lower ones is fully present in this, the simplest case. The other cases are done in the same way.
• The utility of hyperbolic functions in integration can be demonstrated in cases of odd powers of secant (powers of tangent can also be included).
• This is one of several integrals usually done in a first-year calculus course in which the most natural way to proceed involves integrating by parts and returning to the same integral one started with (another is the integral of the product of an exponential function with a sine or cosine function; yet another the integral of a power of the sine or cosine function).
• This integral is used in evaluating any integral of the form

${\displaystyle \int {\sqrt {a^{2}+x^{2}}},円dx,}${\displaystyle \int {\sqrt {a^{2}+x^{2}}},円dx,}

where ${\displaystyle a}${\displaystyle a} is a constant. In particular, it appears in the problems of:

• rectifying the parabola and the Archimedean spiral
• finding the surface area of the helicoid.

This antiderivative may be found by integration by parts, as follows:^[2]

${\displaystyle \int \sec ^{3}x,円dx=\int u,円dv=uv-\int v,円du}${\displaystyle \int \sec ^{3}x,円dx=\int u,円dv=uv-\int v,円du}

where

${\displaystyle u=\sec x,\quad dv=\sec ^{2}x,円dx,\quad v=\tan x,\quad du=\sec x\tan x,円dx.}${\displaystyle u=\sec x,\quad dv=\sec ^{2}x,円dx,\quad v=\tan x,\quad du=\sec x\tan x,円dx.}

Then

${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&=\int (\sec x)(\sec ^{2}x),円dx\\&=\sec x\tan x-\int \tan x,円(\sec x\tan x),円dx\\&=\sec x\tan x-\int \sec x\tan ^{2}x,円dx\\&=\sec x\tan x-\int \sec x,円(\sec ^{2}x-1),円dx\\&=\sec x\tan x-\left(\int \sec ^{3}x,円dx-\int \sec x,円dx\right)\\&=\sec x\tan x-\int \sec ^{3}x,円dx+\int \sec x,円dx.\end{aligned}}}${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&=\int (\sec x)(\sec ^{2}x),円dx\\&=\sec x\tan x-\int \tan x,円(\sec x\tan x),円dx\\&=\sec x\tan x-\int \sec x\tan ^{2}x,円dx\\&=\sec x\tan x-\int \sec x,円(\sec ^{2}x-1),円dx\\&=\sec x\tan x-\left(\int \sec ^{3}x,円dx-\int \sec x,円dx\right)\\&=\sec x\tan x-\int \sec ^{3}x,円dx+\int \sec x,円dx.\end{aligned}}}

Next add ${\textstyle \int \sec ^{3}x,円dx}${\textstyle \int \sec ^{3}x,円dx} to both sides:^[a]

${\displaystyle {\begin{aligned}2\int \sec ^{3}x,円dx&=\sec x\tan x+\int \sec x,円dx\\&=\sec x\tan x+\ln \left|\sec x+\tan x\right|+C,\end{aligned}}}${\displaystyle {\begin{aligned}2\int \sec ^{3}x,円dx&=\sec x\tan x+\int \sec x,円dx\\&=\sec x\tan x+\ln \left|\sec x+\tan x\right|+C,\end{aligned}}}

using the integral of the secant function, ${\textstyle \int \sec x,円dx=\ln \left|\sec x+\tan x\right|+C.}${\textstyle \int \sec x,円dx=\ln \left|\sec x+\tan x\right|+C.}^[2]

Finally, divide both sides by 2:

${\displaystyle \int \sec ^{3}x,円dx={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C,}${\displaystyle \int \sec ^{3}x,円dx={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C,}

which was to be derived.^[2] A possible mnemonic is: "The integral of secant cubed is the average of the derivative and integral of secant".

${\displaystyle \int \sec ^{3}x,円dx=\int {\frac {dx}{\cos ^{3}x}}=\int {\frac {\cos x,円dx}{\cos ^{4}x}}=\int {\frac {\cos x,円dx}{(1-\sin ^{2}x)^{2}}}=\int {\frac {du}{(1-u^{2})^{2}}}}${\displaystyle \int \sec ^{3}x,円dx=\int {\frac {dx}{\cos ^{3}x}}=\int {\frac {\cos x,円dx}{\cos ^{4}x}}=\int {\frac {\cos x,円dx}{(1-\sin ^{2}x)^{2}}}=\int {\frac {du}{(1-u^{2})^{2}}}}

where ${\displaystyle u=\sin x}${\displaystyle u=\sin x}, so that ${\displaystyle du=\cos x,円dx}${\displaystyle du=\cos x,円dx}. This admits a decomposition by partial fractions:

${\displaystyle {\frac {1}{(1-u^{2})^{2}}}={\frac {1}{(1+u)^{2}(1-u)^{2}}}={\frac {1}{4(1+u)}}+{\frac {1}{4(1+u)^{2}}}+{\frac {1}{4(1-u)}}+{\frac {1}{4(1-u)^{2}}}.}${\displaystyle {\frac {1}{(1-u^{2})^{2}}}={\frac {1}{(1+u)^{2}(1-u)^{2}}}={\frac {1}{4(1+u)}}+{\frac {1}{4(1+u)^{2}}}+{\frac {1}{4(1-u)}}+{\frac {1}{4(1-u)^{2}}}.}

Antidifferentiating term-by-term, one gets

${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&={\tfrac {1}{4}}\ln |1+u|-{\frac {1}{4(1+u)}}-{\tfrac {1}{4}}\ln |1-u|+{\frac {1}{4(1-u)}}+C\\[6pt]&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+u}{1-u}}{\Biggl |}+{\frac {u}{2(1-u^{2})}}+C\\[6pt]&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+\sin x}{1-\sin x}}{\Biggl |}+{\frac {\sin x}{2\cos ^{2}x}}+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {1+\sin x}{1-\sin x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{1-\sin ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{\cos ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{2}}\ln \left|{\frac {1+\sin x}{\cos x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{2}}(\ln \left|\sec x+\tan x\right|+\sec x\tan x)+C.\end{aligned}}}${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&={\tfrac {1}{4}}\ln |1+u|-{\frac {1}{4(1+u)}}-{\tfrac {1}{4}}\ln |1-u|+{\frac {1}{4(1-u)}}+C\\[6pt]&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+u}{1-u}}{\Biggl |}+{\frac {u}{2(1-u^{2})}}+C\\[6pt]&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+\sin x}{1-\sin x}}{\Biggl |}+{\frac {\sin x}{2\cos ^{2}x}}+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {1+\sin x}{1-\sin x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{1-\sin ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{\cos ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{2}}\ln \left|{\frac {1+\sin x}{\cos x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\[6pt]&={\tfrac {1}{2}}(\ln \left|\sec x+\tan x\right|+\sec x\tan x)+C.\end{aligned}}}

Alternatively, one may use the tangent half-angle substitution for any rational function of trigonometric functions; for this particular integrand, that method leads to the integration of

${\displaystyle {\frac {2(1+u^{2})^{2}}{(1-u^{2})^{3}}}={\frac {1}{2(1+u)}}-{\frac {1}{2(1+u)^{2}}}+{\frac {1}{(1+u)^{3}}}+{\frac {1}{2(1-u)}}-{\frac {1}{2(1-u)^{2}}}+{\frac {1}{(1-u)^{3}}}.}${\displaystyle {\frac {2(1+u^{2})^{2}}{(1-u^{2})^{3}}}={\frac {1}{2(1+u)}}-{\frac {1}{2(1+u)^{2}}}+{\frac {1}{(1+u)^{3}}}+{\frac {1}{2(1-u)}}-{\frac {1}{2(1-u)^{2}}}+{\frac {1}{(1-u)^{3}}}.}

Integrals of the form: ${\displaystyle \int \sec ^{n}x\tan ^{m}x,円dx}${\displaystyle \int \sec ^{n}x\tan ^{m}x,円dx} can be reduced using the Pythagorean identity if ${\displaystyle n}${\displaystyle n} is even or ${\displaystyle n}${\displaystyle n} and ${\displaystyle m}${\displaystyle m} are both odd. If ${\displaystyle n}${\displaystyle n} is odd and ${\displaystyle m}${\displaystyle m} is even, hyperbolic substitutions can be used to replace the nested integration by parts with hyperbolic power-reducing formulas.

${\displaystyle {\begin{aligned}\sec x&=\cosh u\\[6pt]\tan x&=\sinh u\\[6pt]\sec ^{2}x,円dx&=\cosh u,円du{\text{ or }}\sec x\tan x,円dx=\sinh u,円du\\[6pt]\sec x,円dx&=,円du{\text{ or }}dx=\operatorname {sech} u,円du\\[6pt]u&=\operatorname {arcosh} (\sec x)=\operatorname {arsinh} (\tan x)=\ln |\sec x+\tan x|\end{aligned}}}${\displaystyle {\begin{aligned}\sec x&=\cosh u\\[6pt]\tan x&=\sinh u\\[6pt]\sec ^{2}x,円dx&=\cosh u,円du{\text{ or }}\sec x\tan x,円dx=\sinh u,円du\\[6pt]\sec x,円dx&=,円du{\text{ or }}dx=\operatorname {sech} u,円du\\[6pt]u&=\operatorname {arcosh} (\sec x)=\operatorname {arsinh} (\tan x)=\ln |\sec x+\tan x|\end{aligned}}}

Note that ${\displaystyle \int \sec x,円dx=\ln |\sec x+\tan x|}${\displaystyle \int \sec x,円dx=\ln |\sec x+\tan x|} follows directly from this substitution.

${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&=\int \cosh ^{2}u,円du\\[6pt]&={\tfrac {1}{2}}\int (\cosh 2u+1),円du\\[6pt]&={\tfrac {1}{2}}\left({\tfrac {1}{2}}\sinh 2u+u\right)+C\\[6pt]&={\tfrac {1}{2}}(\sinh u\cosh u+u)+C\\[6pt]&={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C\end{aligned}}}${\displaystyle {\begin{aligned}\int \sec ^{3}x,円dx&=\int \cosh ^{2}u,円du\\[6pt]&={\tfrac {1}{2}}\int (\cosh 2u+1),円du\\[6pt]&={\tfrac {1}{2}}\left({\tfrac {1}{2}}\sinh 2u+u\right)+C\\[6pt]&={\tfrac {1}{2}}(\sinh u\cosh u+u)+C\\[6pt]&={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C\end{aligned}}}

Just as the integration by parts above reduced the integral of secant cubed to the integral of secant to the first power, so a similar process reduces the integral of higher odd powers of secant to lower ones. This is the secant reduction formula, which follows the syntax:

${\displaystyle \int \sec ^{n}x,円dx={\frac {\sec ^{n-2}x\tan x}{n-1}},円+,円{\frac {n-2}{n-1}}\int \sec ^{n-2}x,円dx\qquad {\text{ (for }}n\neq 1{\text{)}},円\!}${\displaystyle \int \sec ^{n}x,円dx={\frac {\sec ^{n-2}x\tan x}{n-1}},円+,円{\frac {n-2}{n-1}}\int \sec ^{n-2}x,円dx\qquad {\text{ (for }}n\neq 1{\text{)}},円\!}

Even powers of tangents can be accommodated by using binomial expansion to form an odd polynomial of secant and using these formulae on the largest term and combining like terms.

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