# Integral of $1/\sinh(x)$

Can you help me find the integral $$\int{\frac{1}{\sinh(x)}dx}?$$

-

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.)

-

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 simpler way to find a primitive of $1/\sin$ is to note that $\mathrm{d}x/\sin(x)$ is invariant by $x\to-x$ hence one knows that the trigonometric line amongst sine, cosine and tangent, invariant by $x\to-x$ yields a change of variables that works. Here $u=\cos(x)$ hence $\mathrm{d}x/\sin(x)=-\mathrm{d}u/(1-u^2)$ and the rest is standard. Likewise, to compute a primitive of $1/\sinh$, use the change of variables $u=\cosh(x)$, then $\mathrm{d}x/\sinh(x)=\mathrm{d}u/(u^2-1)$ and the rest is standard. The point is that this is an automatic way to find a change of variables that works. – Did Apr 5 '11 at 5:55
Likewise, if invariance by $x\to x+\pi$ then use $u=\tan(x)$ and if invariance by $x\to \pi-x$ then use $u=\sin(x)$. These are called les règles de Bioche in French and I do not know the name in English fr.wikipedia.org/wiki/R%C3%A8gles_de_Bioche. – Did Apr 5 '11 at 5:59
And if nothing of the above holds then use $u=\tan(x/2)$. – Did Apr 5 '11 at 6:00
@Didier: The latter is Weierstrass substitution. But I don't know what "the trigonometric line amongst sine, cosine and tangent" means. – Arturo Magidin Apr 5 '11 at 13:50
@Arturo It means that one can use the change of variable $u=\sin(x)$ or the change of variable $u=\cos(x)$ or the change of variable $u=\tan(x)$, depending on the symmetries of the function at hand. And that, if none of these three symmetries hold, one can use the change of variable $u=\tan(x/2)$. The expression règles de Bioche refers to the three first cases. – Did Apr 5 '11 at 14:20

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.

-

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,

-