The integral of hyperbolic cosec I have derived it using basic knowledge of calculus as the following: 
$$K = \int \frac{2}{e^{x}-e^{-x}} \,\mathrm dx = \int \frac{2e^{-x}}{1- e^{-2x}} \, \mathrm dx  $$
using $u$-substitution for $ u = e^{-x } $ then 
$$ K = \int \frac{-2}{1- u^{2}} \, \mathrm du  $$
I don't know if this is valid but I saw that it looked like the derivative of: $ \tanh ^{-1}(x) $ so I made the simplification of taking a factor of $-2$ outside the integeral and proceeded to make $$ K =  -2 \tanh^{-1} (u) + c $$ 

Is that valid and if so, how is it equivelant to the form : $ \ln \left|\tanh \frac{x}{2}\right| + c  $?

 A: $ K = \int \frac{-2}{1- u^{2}} \, \mathrm du  $=$  \int \frac{2}{u^{2}-1} \, \mathrm du=ln|\frac{1-u}{1+u}|  $, backsubbing gives $ln|\frac{1-e^{-x}}{1+e^{-x}}|$ which is $ln|\frac{e^{x}-1}{e^{x}+1}|$. Now $tanhx=\frac{sinhx}{coshx}=\frac{e^x-e^{-x}}{e^x+e^{-x}}=\frac{e^{2x}-1}{e^{2x}+1}$. Now replace $x$ by $x/2$ and put the $ln$ upfront. Can you know reconcile the two? 
A: $$K = \int cosec(x)  \, dx = \log| \tan (x/2)| + c $$
is known circular functions identity where we can use imaginary argument to convert circular trig directly to hyperbolic functions by means of substitution: 
$$ x\rightarrow ix $$
LHS
$$ \int cosec(ix)  \, dix  =\int \frac{dix}{sin(ix)} = =\int \frac{dx}{\sinh(x)}   = \int cosech(x)  \, dx $$
RHS
$$ \log| \tan (ix/2)| + c =  \log| i \tanh (x/2)| + c = \log| \tanh (x/2)| + i \frac{\pi}{2} + c$$
where we evaluated $log\,i$ and so
$$ \int cosech(x)  \, dx  = \log| \tanh (x/2)| + C $$
where the arbitrary constant of integration is now complex and it tallies exactly.
And it affords us an opportunity if you will ( and imho ), seeing several such trig/hyperbolic integrals from a common riemannian foundational viewpoint.
A: You can always check integration results by differentiating.
$$(-2\tanh^{-1}(e^{-x}))'=\frac{2e^{-x}}{1-e^{-2x}}=\frac2{e^x-e^{-x}}=\frac1{\sinh x}$$
and
$$\left(\ln\left|\tanh \frac x2\right|\right)'=\frac{\dfrac1{2\cosh^2 \dfrac x2}}{\tanh \dfrac x2}=\frac1{2\sinh\dfrac x2\cosh\dfrac x2}=\frac1{\sinh x}.$$

For an alternative way to perform the integration,
$$\int\frac{dx}{\sinh x}=\int\frac{\sinh x}{\sinh^2x}dx=\int\frac{\sinh x}{\cosh^2x-1}dx=\int\frac{u}{u^2-1}du=\frac12\log\frac{u-1}{u+1}\\
=\frac12\log\frac{\cosh x-1}{\cosh x+1}.$$
