Integral de la secante cúbica

La integral de la secante cúbica puede ser hallada por el método de integración por partes, en un principio se considera la igualdad:

${\displaystyle \int \sec ^{3}(x)dx=\int \sec(x)\sec ^{2}(x)dx}$

Y se procede por el método de integración por partes considerando que

${\displaystyle {\begin{aligned}u&=\sec x&dv&=\sec ^{2}(x)dx\\du&=\sec(x)\tan(x)dx&v&=\tan(x)dx\end{aligned}}}$

Entonces

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

Si sumamos ${\textstyle \int \sec ^{3}(x)dx}$ a ambos lados de la igualdad y dado que

${\displaystyle \int \sec(x)dx=\ln |\sec(x)+\tan(x)|+C}$

obtenemos

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

Por lo tanto

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

Consideremos que

${\displaystyle {\begin{aligned}\int \sec ^{3}(x)dx&=\int {\frac {dx}{\cos ^{3}(x)}}\\&=\int {\frac {\cos x}{\cos ^{4}(x)}}\;dx\\&=\int {\frac {\cos x}{(1-\sin ^{2}(x))^{2}}}\;dx\\&=\int {\frac {du}{(1-u^{2})^{2}}}\end{aligned}}}$

donde ${\displaystyle u=\operatorname {sen} x}$, de modo que ${\displaystyle du=\cos x\,dx}$.

Esta sustitución admite una descomposición por fracciones parciales como sigue

${\displaystyle {\begin{aligned}{\frac {1}{(1-u^{2})^{2}}}&={\frac {1}{(1+u)^{2}(1-u)^{2}}}\\&={\frac {1/4}{1+u}}+{\frac {1/4}{(1+u)^{2}}}+{\frac {1/4}{1-u}}+{\frac {1/4}{(1-u)^{2}}}\end{aligned}}}$

entonces

${\displaystyle {\begin{aligned}\int \sec ^{3}(x)dx&=\int {\frac {du}{(1-u^{2})^{2}}}\\&=\int \left({\frac {1/4}{1+u}}+{\frac {1/4}{(1+u)^{2}}}+{\frac {1/4}{1-u}}+{\frac {1/4}{(1-u)^{2}}}\right)du\end{aligned}}}$

Si utilizamos linealidad de la integral entonces

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

Integrales de la forma

${\displaystyle \int \sec ^{n}(x)\tan ^{m}(x)dx}$

con ${\displaystyle m,n\in \mathbb {Z} ^{+}}$ pueden ser reducidas utilizando la identidad trigonométrica ${\displaystyle \sec ^{2}(x)-\tan ^{2}(x)=1}$ si ${\displaystyle n}$ es par o ${\displaystyle n}$ y ${\displaystyle m}$ son ambos impares. Si ${\displaystyle n}$ es impar y ${\displaystyle m}$ es par, las sustituciones hiperbólicas suelen ser usadas para evitar el uso del método de integración por partes, para así sólo reducir potencias de funciones hiperbólicas.

Dado que

${\displaystyle {\begin{aligned}\sec(x)&=\cosh(u)\\[6pt]\tan(x)&=\sinh(u)\\[6pt]\end{aligned}}}$

entonces

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

Nótese que ${\textstyle \int \sec x\,dx=\ln |\sec x+\tan x|}$ se sigue directamente de esta sustitución.

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

Se desea calcular

${\displaystyle \int \sec ^{5}(x)dx}$

Comencemos considerando que

${\displaystyle \int \sec ^{5}(x)dx=\int \sec ^{3}(x)\sec ^{2}(x)dx}$

Y procedemos por el método de integración por partes considerando que

${\displaystyle {\begin{aligned}u&=\sec ^{3}(x)&dv&=\sec ^{2}(x)dx\\du&=3\sec ^{3}(x)\tan(x)dx&v&=\tan(x)dx\end{aligned}}}$

Entonces

${\displaystyle {\begin{aligned}\int \sec ^{5}(x)dx&=\int \sec ^{3}(x)\sec ^{2}(x)dx\\&=\sec ^{3}(x)\tan(x)-3\int \sec ^{3}(x)\tan ^{2}(x)dx\\&=\sec ^{3}(x)\tan(x)-3\int \sec ^{3}(x)\left[\sec ^{2}(x)-1\right]dx\\&=\sec ^{3}(x)\tan(x)-3\int \sec ^{5}(x)dx+3\int \sec ^{3}(x)dx\\4\int \sec ^{5}(x)dx&=\sec ^{3}(x)\tan(x)+3\left({\frac {\sec(x)\tan(x)+\ln |\sec(x)+\tan(x)|}{2}}\right)+C\\\int \sec ^{5}(x)dx&={\frac {\sec ^{3}(x)\tan(x)}{4}}+{\frac {3(\sec(x)\tan(x)+\ln |\sec(x)+\tan(x)|)}{8}}+C\end{aligned}}}$

Uno puede demostrar utilizando integración por partes que la fórmula de reducción para la función secante está dada por:

${\displaystyle \int \sec ^{n}(x)dx={\frac {\sec ^{n-2}(x)\tan(x)}{n-1}}+{\frac {n-2}{n-1}}\int \sec ^{n-2}(x)dx}$

para ${\displaystyle n\geq 2}$ o alternativamente

${\displaystyle \int \sec ^{n}(x)dx={\frac {\sec ^{n-1}(x)\operatorname {sen}(x)}{n-1}}+{\frac {n-2}{n-1}}\int \sec ^{n-2}(x)dx}$