# How do you derive the formula of S csch(x)(dx)=ln (tanh (x/2)) +C ?

• Mar 6th 2009, 03:41 AM
How do you derive the formula of S csch(x)(dx)=ln (tanh (x/2)) +C ?
The integrand can be transformed to functions of tanh (x/2) by first observing that sinh 2A = 2sinh A cosh A,

from which

csch(x) = 1/(2sinh(x/2)cosh(x/2)).

Clever algebra, followed by an application of the Pythagorean properties, produces the desired result. Then substitute u for tanh(x/2). You will have to be clever again to figure out what to substitute for dx in terms of du. The result integral is remarkably simple! Confirm that the formula works by evaluatinng the integral on the interval (1,2), then checking by numerical integration.

This has got me confused!!!
• Mar 6th 2009, 04:51 AM
Chop Suey
Use the identities:
$\displaystyle 1+\cosh{x} = 2\cosh^2{\frac{x}{2}}$
$\displaystyle \sinh{x} = 2\sinh{\frac{x}{2}}\cosh{\frac{x}{2}}$

Apply a "clever algebra" trick by multiplying the integrand with

$\displaystyle \frac{\text{csch}{x}+\coth{x}}{\text{csch}{x}+\cot h{x}}$

to get

$\displaystyle \int \frac{\text{csch}^2{x}+\coth{x}\text{csch}{x}}{\te xt{csch}{x}+\coth{x}}~dx$

Note that the integral is simply

$\displaystyle -\ln{|\text{csch}{x}+\coth{x}|} + C$

which is actually

$\displaystyle \ln{\left|\frac{\sinh{x}}{\cosh{x}+1}\right|} + C$

Apply the aforementioned identities to get:

$\displaystyle \ln{\left|\frac{\sinh{x}}{\cosh{x}+1}\right|} = \ln{\left|\frac{2\sinh{\frac{x}{2}\cosh{\frac{x}{2 }}}}{2\cosh^2{\frac{x}{2}}}\right|} = \ldots$

EDIT: Perhaps the question was looking for a different approach. Knowing that $\displaystyle \cosh^2{x}-\sinh^2{x} = 1$, we can switch the integrand to:
$\displaystyle \frac{1}{\sinh{x}} = \frac{\cosh^2{\frac{x}{2}}-\sinh^2{\frac{x}{2}}}{2\sinh{\frac{x}{2}}\cosh{\fr ac{x}{2}}} = \frac{\cosh{\frac{x}{2}}}{2\sinh{\frac{x}{2}}}-\frac{\sinh{\frac{x}{2}}}{2\cosh{\frac{x}{2}}}$

Which are two straightforward integrals. After evaluating the integral, apply the logarithm law $\displaystyle \ln{A} - \ln{B} = \ln{\left(\frac{A}{B}\right)}$
• Mar 6th 2009, 05:36 AM