Integrate sec x
(A)
Yue Kwok Choy
There are different solutions to integrate sec x. Readers are suggested to show that they are
equivalent.
)*
(1) sec x dx = =
+,- *
=
?
1 7
-@+20 3 4)*
2
6 2
1 7
AB/0 3 4
6 2
=
)*
)*
1
2
-./0 3*4
1 7
6 2
1 7
AB/0 3 4
6 2
)CAB/0 3 4D
(2) sec x dx = +,- * = +,- *)*
+,-2 *
=
)*
1 7
6 2
1 7
6 2
5-./0 3 4 +,-0 3 4
J
8
= 9:;01<74
5
)*
6 2 +,-201374
1 7
6 2
=>90 < 4
6 2
= EF GHIF 0K + M4G + +,- *)*
= (83-./ *)(8O-./ *)
8
8
8
8
8
Put t = sin x, sec x dx = (83A)(8OA) dt = 5 C83A + 8OAD dt = 5 Rln|1 + t| − ln|1 − t|U + c
8
83A
V
V3WF = 5 ln G8OAG + c = M EF GVOWF G + (3) sec x dx = -@+ *(-@+ *3AB/ *)
-@+ *3AB/ *
dx = Y-@+ * AB/ *3 -@+2 *Z)*
-@+ *3AB/ *
=
)(-@+ *3AB/ *)
-@+ *3AB/ *
= EF| + HIF | + (4) Put sec x = cosh θ , tan x = √sec 5 θ − 1 = √cosh5 θ − 1 = sinh θ
Also, we can get sin x = tanh θ (please check yourselves)
Therefore sec 5 x dx = cosh θ dθ ⟹ dx =
+,-` a)a
-@+2 *
=
+,-` a)a
+,-`2 a
)a
= +,-` a
)a
sec x dx = cosh θ +,-` a = dθ = θ + c = coshO8 (sec x) + c
∴ sec x dx = cdOV ( ) + = WFdOV (HIF ) + = HIFdOV (WF ) + )-
)-
(5) Put s = sec x , ds = sec x tan x dx, sec x dx = AB/ * = √-2
O8
)-
sec x dx = √-2 O8 = lnfs + √s 5 − 1f + c = EF| + HIF | + )-
Proof of: √-2 = lnfs + √s5 − 1f + c
O8
Put u5 = s5 − 1, 2udu = 2sds ⟹
)-
√-2 O8 = )h
=
)(-3h)
-3h
du ds du + ds d(s + u)
=
=
=
s
u
s+u
s+u
= ln|s + u| + c = lnfs + √s5 − 1f + c
(6) Put
*
8OA2
5)A
83A2
t = tan 5 , dx = 83A2 , cos x = 83A2 , sec x = 8OA2
83A2
5)A
5)A
8
8
83A
M
VOHIF
M
V3HIF
sec x dx = 8OA2 C83A2 D = 8OA2 = C83A + 8OAD dt = ln G8OAG + c = EF j
*
Also, putting tan =
5
7
2
7
+,2
-./
, we have
M
M
c OWF
M
M
c 3WF
sec x dx = EF j
j+
j+
(B) F for different natural numbers n.
(1) sec 5 x dx = tan x + c
(2) I = sec n x dx = sec x sec 5 x dx = sec x d(tan x)
= sec xtan x − tan xd(sec x) (integation by parts)
= sec xtan x − tan x sec x tan x dx
= sec xtan x − tan5 x sec x dx = sec xtan x − (sec 5 x − 1) sec x dx
= sec xtan x − sec n x dx + sec x dx
= sec xtan x − I + ln|sec x + tan x|
2I = sec xtan x + ln|sec x + tan x|
V
I = sec n x dx = M R HIF + EF| + HIF |U + (3) sec o x dx = (1 + tan5 x) sec 5 x dx = (1 + tan5 x)d(tan x) = HIF +
HIFp p
(4) Reduction formula for F (used mostly for n is odd)
I/ = sec / x dx = sec /O5 x d(tan x) = sec /O5 x tan x − tan x d(sec /O5 x)
= sec /O5 x tan x − tan x (n − 2) sec /On x (sec x tan x)dx
= sec /O5 x tan x − (n − 2) tan5 x sec /O5 x dx
= sec /O5 x tan x − (n − 2) (sec 5 x − 1) sec /O5 x dx
= sec /O5 x tan x − (n − 2)I/ + (n − 2)I/O5
V
qF = FOV R FOM HIF + (F − M)qFOM U
8
n
8
n 8
(5) Ir = sec r x dx = o sec n x tan x + o In = o sec n x tan x + o C5 (sec x tan x + I8 )D
V
p
= K p HIF + s R HIF + EF| + HIF |U + +
(6) F (n = 2m is even)
sec 5t x dx = sec 5tO5 x d(tan x) = (1 + tan5 x)tO8 d(tan x)
= (1 + u5 )tO8 du (where u = tan x)
tO8 5v
tO8
tO8
= ∑tO8
u du = ∑tO8
u5v du = ∑tO8
vwx Cv
vwx Cv
vwx Cv
zOV
Mz = ∑zOV
|w} {|
(7) sec  x dx = ∑5vwx Cv5
(AB/ *)2y<?
5v38
M
(HIF )M|<V
M|3V
h2y<?
5v38
+c
+
= Cx5 tan x + C85
(AB/ *)€
n
+ C55
(AB/ *)
r
+c
M
= tIF + (HIF )p + (HIF )‚ + p
‚
(C) √ (for interest)
It is integrable, but cannot be expressed in terms of elementary functions. It involves elliptic
integral of the first kind.
√ = M ƒ 0M , M4 , where F(x, m) is the elliptic integral of the first kind with parameter
m = k5 .
Maclaurin expansion of √sec x ≈ 1 +
√sec xdx ≈ 01 +
*2
o
+
‡ *6
ˆ
+
8nˆ *‰
r‡x
*2
o
+
‡ *6
ˆ
+
4 dx ≈ +
8nˆ *‰
r‡x
p
VM
+
+⋯
‹ ‚
Ks}
+
VpŒ ‹
K}pM}
+