Integral csch(x)

Themes > Science > Mathematics > Calculus > Integrals > Integrals Table > Integral 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 =

sinh x

e^x - e^-x

csch x dx = (integral)

e^x - e^-x

multiply numerator and denominator by e^x

(e^x - e^-x)

(e^x)

2 e^x

(e^2x - 1)

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

substitute du = e^x dx, u = e^x

= (integral) 2 du
u² - 1

the denominator is factorable; use partial fractions

= 2 du
(u + 1)(u - 1)
= A
u + 1
+ B
u - 1
du

(u + 1)(u - 1)

u + 1

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

= -

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 | e^x - 1
e^x + 1
| + C
multiply the numerator and denominator by e^-x/2

= ln | (e^x - 1) e^-x/2
(e^x + 1) e^-x/2
| + C = ln | e^x/2 - e^-x/2
e^x/2 + e^-x/2
| + C

recall that

tanh x =

e^x - e^-x

e^x + e^-x

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

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