Commonly encountered and tricky integral
The integral of secant cubed is a frequent and challengingindefinite 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\\[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
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 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 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
∫
a
2
+
x
2
d
x
,
{\displaystyle \int {\sqrt {a^{2}+x^{2}}}\,dx,}
where
a
{\displaystyle a}
is a constant. In particular, it appears in the problems of:
Derivations
Integration by parts
This antiderivative may be found by integration by parts , as follows:[2]
∫
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:[a]
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.}
[2]
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.[2] A possible mnemonic is: "The integral of secant cubed is the average of the derivative and integral of secant".
Reduction to an integral of a rational function
∫
sec
3
x
d
x
=
∫
d
x
cos
3
x
=
∫
cos
x
d
x
cos
4
x
=
∫
cos
x
d
x
(
1
−
sin
2
x
)
2
=
∫
d
u
(
1
−
u
2
)
2
{\displaystyle \int \sec ^{3}x\,dx=\int {\frac {dx}{\cos ^{3}x}}=\int {\frac {\cos x\,dx}{\cos ^{4}x}}=\int {\frac {\cos x\,dx}{(1-\sin ^{2}x)^{2}}}=\int {\frac {du}{(1-u^{2})^{2}}}}
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\\[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 |\sec x+\tan x|+\sec x\tan x)+C.\end{aligned}}}
Hyperbolic functions
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\\[6pt]\tan x&=\sinh u\\[6pt]\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}}}
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\\[6pt]&={\tfrac {1}{2}}\int (\cosh 2u+1)\,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}}}
Higher odd powers of secant
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
of secant and using these formulae on the largest term and combining like terms.
See also
Notes
constants of integration
are absorbed in the remaining integral term.
References
^ Spivak, Michael (2008). "Integration in Elementary Terms". Calculus . p. 382 . This is a tricky and important integral that often comes up.
^ .