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The integral of secant cubed is a frequent and challenging indefinite integral of elementary calculus:
∫ sec 3 x d x = 1 2 sec x tan x + 1 2 ∫ sec x d x + C = 1 2 ( sec x tan x + ln | sec x + tan x | ) + C = 1 2 ( sec x tan x + gd − 1 x ) + C , | x | < 1 2 π {\textstyle {\begin{aligned}\int \sec ^{3}x\,dx&={\tfrac {1}{2}}\sec x\tan x+{\tfrac {1}{2}}\int \sec x\,dx+C\\&={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C\\&={\tfrac {1}{2}}(\sec x\tan x+\operatorname {gd} ^{-1}x)+C,\qquad |x|<{\tfrac {1}{2}}\pi \end{aligned}}}where 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:
This antiderivative may be found by integration by parts, as follows:
∫ sec 3 x d x = ∫ u d v = u v − ∫ v d u {\displaystyle \int \sec ^{3}x\,dx=\int u\,dv=uv-\int v\,du}where
u = sec x , d v = sec 2 x d x , v = tan x , d u = sec x tan x d x . {\displaystyle u=\sec x,\quad dv=\sec ^{2}x\,dx,\quad v=\tan x,\quad du=\sec x\tan x\,dx.}Then
∫ sec 3 x d x = ∫ ( sec x ) ( sec 2 x ) d x = sec x tan x − ∫ tan x ( sec x tan x ) d x = sec x tan x − ∫ sec x tan 2 x d x = sec x tan x − ∫ sec x ( sec 2 x − 1 ) d x = sec x tan x − ( ∫ sec 3 x d x − ∫ sec x d x ) = sec x tan x − ∫ sec 3 x d x + ∫ sec x d x . {\displaystyle {\begin{aligned}\int \sec ^{3}x\,dx&=\int (\sec x)(\sec ^{2}x)\,dx\\&=\sec x\tan x-\int \tan x\,(\sec x\tan x)\,dx\\&=\sec x\tan x-\int \sec x\tan ^{2}x\,dx\\&=\sec x\tan x-\int \sec x\,(\sec ^{2}x-1)\,dx\\&=\sec x\tan x-\left(\int \sec ^{3}x\,dx-\int \sec x\,dx\right)\\&=\sec x\tan x-\int \sec ^{3}x\,dx+\int \sec x\,dx.\end{aligned}}}Next add ∫ sec 3 x d x {\textstyle \int \sec ^{3}x\,dx} to both sides:
2 ∫ sec 3 x d x = sec x tan x + ∫ sec x d x = sec x tan x + ln | sec x + tan x | + C , {\displaystyle {\begin{aligned}2\int \sec ^{3}x\,dx&=\sec x\tan x+\int \sec x\,dx\\&=\sec x\tan x+\ln \left|\sec x+\tan x\right|+C,\end{aligned}}}using the integral of the secant function, ∫ sec x d x = ln | sec x + tan x | + C . {\textstyle \int \sec x\,dx=\ln \left|\sec x+\tan x\right|+C.}
Finally, divide both sides by 2:
∫ sec 3 x d x = 1 2 ( sec x tan x + ln | sec x + tan x | ) + C , {\displaystyle \int \sec ^{3}x\,dx={\tfrac {1}{2}}(\sec x\tan x+\ln \left|\sec x+\tan x\right|)+C,}which was to be derived. A possible mnemonic is: "The integral of secant cubed is the average of the derivative and integral of secant".
where u = sin x {\displaystyle u=\sin x} , so that d u = cos x d x {\displaystyle du=\cos x\,dx} . This admits a decomposition by partial fractions:
1 ( 1 − u 2 ) 2 = 1 ( 1 + u ) 2 ( 1 − u ) 2 = 1 4 ( 1 + u ) + 1 4 ( 1 + u ) 2 + 1 4 ( 1 − u ) + 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
∫ sec 3 x d x = 1 4 ln | 1 + u | − 1 4 ( 1 + u ) − 1 4 ln | 1 − u | + 1 4 ( 1 − u ) + C = 1 4 ln | 1 + u 1 − u | + u 2 ( 1 − u 2 ) + C = 1 4 ln | 1 + sin x 1 − sin x | + sin x 2 cos 2 x + C = 1 4 ln | 1 + sin x 1 − sin x | + 1 2 sec x tan x + C = 1 4 ln | ( 1 + sin x ) 2 1 − sin 2 x | + 1 2 sec x tan x + C = 1 4 ln | ( 1 + sin x ) 2 cos 2 x | + 1 2 sec x tan x + C = 1 2 ln | 1 + sin x cos x | + 1 2 sec x tan x + C = 1 2 ( ln | sec x + tan x | + sec x tan x ) + C . {\displaystyle {\begin{aligned}\int \sec ^{3}x\,dx&={\tfrac {1}{4}}\ln |1+u|-{\frac {1}{4(1+u)}}-{\tfrac {1}{4}}\ln |1-u|+{\frac {1}{4(1-u)}}+C\\&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+u}{1-u}}{\Biggl |}+{\frac {u}{2(1-u^{2})}}+C\\&={\tfrac {1}{4}}\ln {\Biggl |}{\frac {1+\sin x}{1-\sin x}}{\Biggl |}+{\frac {\sin x}{2\cos ^{2}x}}+C\\&={\tfrac {1}{4}}\ln \left|{\frac {1+\sin x}{1-\sin x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{1-\sin ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\&={\tfrac {1}{4}}\ln \left|{\frac {(1+\sin x)^{2}}{\cos ^{2}x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\&={\tfrac {1}{2}}\ln \left|{\frac {1+\sin x}{\cos x}}\right|+{\tfrac {1}{2}}\sec x\tan x+C\\&={\tfrac {1}{2}}(\ln |\sec x+\tan x|+\sec x\tan x)+C.\end{aligned}}}Integrals of the form: ∫ sec n x tan m x d x {\displaystyle \int \sec ^{n}x\tan ^{m}x\,dx} can be reduced using the Pythagorean identity if n {\displaystyle n} is even or n {\displaystyle n} and m {\displaystyle m} are both odd. If n {\displaystyle n} is odd and m {\displaystyle m} is even, hyperbolic substitutions can be used to replace the nested integration by parts with hyperbolic power-reducing formulas.
sec x = cosh u tan x = sinh u sec 2 x d x = cosh u d u or sec x tan x d x = sinh u d u sec x d x = d u or d x = sech u d u u = arcosh ( sec x ) = arsinh ( tan x ) = ln | sec x + tan x | {\displaystyle {\begin{aligned}\sec x&=\cosh u\\\tan x&=\sinh u\\\sec ^{2}x\,dx&=\cosh u\,du{\text{ or }}\sec x\tan x\,dx=\sinh u\,du\\\sec x\,dx&=\,du{\text{ or }}dx=\operatorname {sech} u\,du\\u&=\operatorname {arcosh} (\sec x)=\operatorname {arsinh} (\tan x)=\ln |\sec x+\tan x|\end{aligned}}}Note that ∫ sec x d x = ln | sec x + tan x | {\displaystyle \int \sec x\,dx=\ln |\sec x+\tan x|} follows directly from this substitution.
∫ sec 3 x d x = ∫ cosh 2 u d u = 1 2 ∫ ( cosh 2 u + 1 ) d u = 1 2 ( 1 2 sinh 2 u + u ) + C = 1 2 ( sinh u cosh u + u ) + C = 1 2 ( sec x tan x + ln | sec x + tan x | ) + C {\displaystyle {\begin{aligned}\int \sec ^{3}x\,dx&=\int \cosh ^{2}u\,du\\&={\tfrac {1}{2}}\int (\cosh 2u+1)\,du\\&={\tfrac {1}{2}}\left({\tfrac {1}{2}}\sinh 2u+u\right)+C\\&={\tfrac {1}{2}}(\sinh u\cosh u+u)+C\\&={\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:
∫ sec n x d x = sec n − 2 x tan x n − 1 + n − 2 n − 1 ∫ sec n − 2 x d x (for n ≠ 1 ) {\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\qquad {\text{ (for }}n\neq 1{\text{)}}\,\!}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.