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 ex  = 2  (ex - e-x) (ex)  (ex) dx
 = 2 ex  (e2x - 1) dx
set
u = ex
then we find
du = ex dx

substitute du = ex dx, u = ex
 = 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
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 = ex  = 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.