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Proof: Integral csch(x)
(Math | Calculus | Integrals | Table Of | csch x)
 
Discussion of
(integral) 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
     
    (integral) csch x dx = (integral)
    2 
    ex - ex
    		dx
    multiply numerator and denominator by ex
    = (integral) 
    2 
    (ex - ex)
    		(ex) 
    (ex)    		 dx
     
     = (integral)
    2 ex 
    (e2x - 1)
    		dx
    set
      u = ex
    then we find
      du = ex dx

    substitute du = ex dx, u = ex
     
    = (integral)
    2 du 
    u2 - 1
    the denominator is factorable; use partial fractions
    = (integral)
    2 du
    (u + 1)(u - 1)
    = (integral)
    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

     
    = (integral)
    -1 du 
    u + 1
    		+ (integral)
    1 du 
    u - 1
    set v = u + 1, w = u - 1
    then we find dv = du, dw = du
    substitute
    = - (integral)
    dv 
    v
    		+ (integral)
    dw 
    w
    solve both integrals

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

    substitute 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 ex/2
    = ln |
    (ex - 1) ex/2 
    (ex + 1) ex/2
    		| + C = ln | ex/2 - ex/2 
    ex/2 + ex/2    		 | + C

    recall that

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

 
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