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.
Best Answer
With $t=\sinh x$, i.e. $x=\text{arsinh }t$,
$$\int\frac{dx}{\sqrt{1-\text{csch}^2x}}=\int\frac{dt}{\sqrt{t^2+1}\sqrt{1-\dfrac1{t^2}}}=\int\frac{t\,dt}{\sqrt{t^4-1}}=\frac12\int\frac{\,dt^2}{\sqrt{(t^2)^2-1}}=\frac12\text{arcosh }t^2.$$