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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 $\sinh(x)$ signal, 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.

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  • $\begingroup$ Please show what you have tried $\endgroup$ – clathratus Feb 4 at 20:29
  • $\begingroup$ @clathratus Ok! $\endgroup$ – Élio Pereira Feb 4 at 20:32
  • $\begingroup$ Note that $$1-\frac1{\sinh(x)^2}=\frac{\sinh(x)^2-1}{\sinh(x)^2}$$ So $$\frac1{\sqrt{1-\text{csch}(x)^2}}=\frac{\sinh(x)}{\sqrt{\sinh(x)^2-1}}$$ $\endgroup$ – clathratus Feb 4 at 20:41
  • $\begingroup$ Shouldn't it be $|\sinh(x)|$ in the numerator instead? $\endgroup$ – Élio Pereira Feb 4 at 20:44
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    $\begingroup$ Thank you for editing your question! (+1) $\endgroup$ – clathratus Feb 4 at 20:44
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With $t=\sinh x$, i.e. $x=\text{arsinh }t$,

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

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    $\begingroup$ No, that would be $\frac{dt}{\sqrt{t^2+1}}$. It doesn't work. $\endgroup$ – jmerry Feb 4 at 20:57
  • $\begingroup$ @jmerry: fixed since. $\endgroup$ – Yves Daoust Feb 4 at 20:59
  • $\begingroup$ @jmerry Thank you for your help. I will now study the problem related to $x$ signal. $\endgroup$ – Élio Pereira Feb 4 at 22:11
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    $\begingroup$ @Élio Pereira: Uh - not my answer. I spotted an error that was then quickly corrected, but that's all. $\endgroup$ – jmerry Feb 4 at 22:19
  • $\begingroup$ @YvesDaoust Thank you for help. (@jmerry I made a mistake, my intention was to thank Yves :), but anyway you should also receive some credit.) $\endgroup$ – Élio Pereira Feb 4 at 22:29
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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}

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