Integral of $1/\sinh(x)$ Can you help me find the integral 
$$
\int{\frac{1}{\sinh(x)}dx}?
$$
 A: In addition to Adrián Barquero's suggestion, another idea is to exploit the similarity between hyperbolic and trigonometric functions.
Namely, this integral can be solved using the same kind of idea as its trigonometric counterpart.
To solve $\displaystyle \int\frac{1}{\sin x}\,dx = \int \csc x\,dx$, we multiply and divide by $\csc x + \cot x$, then do the substitution $u = \csc x + \cot x$, with $du = (-\csc x\cot x - \csc^2 x)\,dx$. We have:
$$\begin{align*}
\int\csc x\,dx &= \int \frac{\csc x(\csc x + \cot x)}{\csc x + \cot x}\,dx\\
&= \int\frac{\csc^2 x + \csc x\cot x}{\csc x+\cot x}\,dx\\
&= \int \frac{-du}{u} = -\ln|u| +C\\
&= -\ln|\csc x + \cot x| + C.
\end{align*}$$
The same idea works for hyperbolic functions, if you use the fact that $(\coth x)' = -\mathrm{csch}^2 x$ and $(\mathrm{csch} x)' = -\mathrm{csch} x\coth x$. So write
$$\int\frac{1}{\sinh x}\,dx = \int \mathrm{csch} x\,dx$$
and multiply and divide by $\mathrm{csch} x + \coth x$ in preparation for a change of variable.
A: Hint:   Notice that since $\sinh(x)=2\sinh(x/2)\cosh(x/2)$ and $1=\cosh^2(x/2)-\sinh^2(x/2)$ we can rewrite our integrand as $$\frac{1}{\sinh(x)}=\frac{\cosh^2(x/2)-\sinh^2(x/2)}{2\sinh(x/2)\cosh(x/2)}=\frac{\cosh(x/2)}{\sinh(x/2)}-\frac{\sinh(x/2)}{\cosh(x/2)}.$$
Can you solve it from here?  
Hint 2: What is the derivative of $\log (\sinh(x))$?
Hope that helps,
A: Arturo's answer brought to mind yet another way of computing this integral: cheat by using the formula $$\sinh x = -i \sin ix,$$ where $i = \sqrt{-1}$.  You can manipulate the imaginary units like ordinary (real) constants for the purposes of integration, and the result follows from the formula for $\int \csc x\,dx.$
Admittedly, the answer you'll get will have imaginary units in them, but you should be able to get rid of those via similar formulas (such as $\cos ix = \cosh x$, for example.)
A: If you write the integral as 
$$\int \frac{1}{\sinh{x}} \, dx = \int \frac{2 e^x}{e^{2x} - 1} \, dx$$
then there's a pretty obvious substitution that works.
