Integrating Hyperbolic Functions

Recall the de?nitions of the hyperbolic cosine and hyperbolic sine functions as
.
cosh x =
ex + e??x
2
and sinh x =
ex ?? e??x
2
sech x =
1
cosh x
and csch x =
1
sinh x
tanh x =
sinh x
cosh x
and coth x =
cosh x
sinh x
and also recall that
cosh2 x ?? sinh2 x = 1 for all real number x.
Also note that when solving for one in terms of the other, cosh2 x and sinh2 x behave slightly di¤erently. Alge-
braically, cosh2 x = 
p
1 + sinh2 x and sinh2 x = 
p
cosh2 x ?? 1. However, cosh x is always positive (as a matter
of fact, always at least 1) and so
cosh x =
p
1 + sinh2 x
while sinh x is negative for negative x and so
sinh x = 
p
cosh2 x ?? 1 =
( p
cosh2 x ?? 1 if x  0
??
p
cosh2 x ?? 1 if x < 0
It is easy to verify that
d
dx
(sinh x) = cosh x and
d
dx
(cosh x) = sinh x. Therefore,
Theorem 1:
Z
sinh xdx = cosh x + C and
Z
cosh xdx = sinh x + C
Theorem 2:
Z
tanh xdx = ln (cosh x) + C and
Z
coth xdx = ln jsinh xj + C
proof: We will compute
Z
tanh xdx.
We will use substitution. Let u = cosh x: Then du = sinh xdx and so
Z
tanh xdx =
Z
sinh x
cosh x
dx =
Z
1
cosh x
(sinh xdx) =
Z
1
u
du = ln juj + C = ln jcosh xj + C = ln (cosh x) + C
The last step is because cosh x is always positive. The other integral,
Z
coth xdx goes very similarly, using the
substitution u = sinh x.
Theorem 3:
Z
sech xdx = 2 tan??1 (ex) + C
proof:
Z
sech xdx =
Z
1
cosh x
dx =
Z
2
ex + e??x dx =
Z
2
ex + e??x

ex
ex dx =
Z
2ex
(ex)2 + e??x (ex)
dx
=
Z
2ex
(ex)2 + e??x+x
dx =
Z
2ex
(ex)2 + e0
dx =
Z
2ex
(ex)2 + 1
dx