Indefinite integral $\int\frac{1}{\sqrt{1-\text{csch}^2 x}}\,dx$

Question

I would like to know how to solve the following indefinite integral:

$$I=\int\frac{1}{\sqrt{1-\text{csch}^2(x)}}\,dx$$

where csch$(x)$ is the the hyperbolic cosecant function of $x$, i.e. csch$(x)=\frac{1}{\sinh(x)}$.

I tried the substitution $u=\sqrt{1-\text{csch}^2(x)}$, which lead to $I=\int \frac{\sinh^3(x)}{\cosh(x)}\,du$. From the substitution relation I got:

$$\sinh(x)=\pm\frac{1}{\sqrt{1-u^2}},\hspace{50pt}\cosh(x)=\frac{\sqrt{2-u^2}}{\sqrt{1-u^2}}.$$

There's the problem with the sign of $\sinh(x)$. Since in the specific problem where I found this integral, $x$ is a function of $t$, $x(t)$ and this function isn't determined. Also, the substitution didn't simplify enough the problem. I also tried the substitution $u=\text{csch}(x)$, but I got similar problems.

Answer 1

With $t=\sinh x$, i.e. $x=\text{arsinh }t$,

$$\int\frac{dx}{\sqrt{1-\text{csch}^2x}}=\int\frac{dt}{\sqrt{t^2+1}\sqrt{1-\dfrac1{t^2}}}=\int\frac{t\,dt}{\sqrt{t^4-1}}=\frac12\int\frac{\,dt^2}{\sqrt{(t^2)^2-1}}=\frac12\text{arcosh }t^2.$$

Answer 2

Here is an alternative method.

Since $\operatorname{csch} x = 1/\sinh x$ we can write \begin{align} \int \frac{dx}{\sqrt{1 - \operatorname{csch}^2 x}} &= \int \frac{\sinh x}{\sqrt{\sinh^2 x - 1}} \, dx, \qquad x > 0\\ &= \int \frac{\sinh x}{\sqrt{\cosh^2 - 2}} \, dx \end{align} where we have used $\cosh^2 x - \sinh^2 = 1$. Setting $t = \cosh x$, $dt = \sinh x, dx$ one has \begin{align} \int \frac{dx}{\sqrt{1 - \operatorname{csch}^2 x}} &= \int \frac{dt}{\sqrt{t^2 - 2}}\\ &= \cosh^{-1} \left (\frac{t}{\sqrt{2}} \right ) + C\\ &= \cosh^{-1} \left (\frac{\cosh x}{\sqrt{2}} \right ) + C, \qquad x > 0. \end{align}

Answer 3

To avoid sign complications, substitute $t= \coth x$ instead

\begin{align} &\int\frac{1}{\sqrt{1-\text{csch}^2 x}}\,dx\\ =&\int\frac{dt}{(1-t^2)\sqrt{2-t^2}} = \tanh^{-1}\frac{\sqrt{2-t^2}}{t} = \tanh^{-1}\frac{\sqrt{1-\text{csch}^2x}}{\coth x} \end{align} which is valid for all domain of $x$.