Dave's Math Tables: Integral csch(x)

(Math | Calculus | Integrals | Table Of | csch x)

Discussion of csch x dx = ln | tanh(x/2) | + C.

1. Proof

Strategy: Use definition of csch; use algebra; use substitution; use partial fractions.

csch x =

1  sinh x

=

2  ex - e-x

 

csch x dx =

2  ex - e-x

dx

multiply numerator and denominator by e^x

=

2  (ex - e-x)

(ex)  (ex)

dx

 

=

2 ex  (e2x - 1)

dx

set
  u = e^x
then we find
  du = e^x dx

substitute du = e^x dx, u = e^x
 

=	2 du  u2 - 1

the denominator is factorable; use partial fractions

=

2 du (u + 1)(u - 1)

=

A  u + 1

+

B  u - 1

du

 

2  (u + 1)(u - 1)

=

A  u + 1

+

B  u - 1

multiply both sides by (u + 1)(u - 1)

2 = A(u - 1) + B(u + 1)
2 = Au - A + Bu + B
2 + 0u = (-A + B) + (A + B)u

therefore
-A + B = 2
 A + B = 0
------------- (add)
       2B = 2
         B = 1
A + 1 = 0
A = -1

 

=

-1 du  u + 1

+

1 du  u - 1

set v = u + 1, w = u - 1
then we find dv = du, dw = du
substitute

= -

dv  v

+

dw  w

solve both integrals

= - ln |v| + ln |w| + C

sustitute back v = u + 1, w = u - 1

= - ln |u + 1| + ln | u - 1| + C

= ln |

u - 1  u + 1

| + C

substitute back u = e^x

= ln |

ex - 1  ex + 1

| + C

multiply the numerator and denominator by e^-x/2

= ln |

(ex - 1) e-x/2  (ex + 1) e-x/2

| + C

= ln |

ex/2 - e-x/2  ex/2 + e-x/2

| + C


recall that

tanh x =

ex - e-x ex + e-x

therefore
= ln | tanh(x/2) | + C
Q.E.D.