Hyperbolic substitution for $\int\frac{dx}{x\sqrt{1-x^2}}$

Question

Substitute $x=\sinh\theta$, $\cosh\theta$ or $\tanh\theta$. After integration change back to $x$.$$\int\frac{dx}{x\sqrt{1-x^2}}$$

Substituting $x=\tanh\theta$, we have $$\begin{align} \int\frac{dx}{x\sqrt{1-x^2}}&=\int\frac{\text{sech}^2\,\theta\,d\theta}{\tanh\theta\sqrt{1-\tanh^2\theta}}\\ &=\int\text{csch }\theta\,d\theta\\ &=-\ln|\text{csch }\theta+\text{coth }\theta|+C \end{align}$$ Here comes the problem. We can substitute back $\text{coth }\theta=\frac1x$ but what do we have for $\text{csch }\theta$? We have $\text{csch}^2\,\theta=\coth^2\theta-1$, but neither $\sqrt{\coth^2\theta-1}$ nor $-\sqrt{\coth^2\theta-1}$ seems to be a one-off solution for $\text{csch }\theta$. How should we proceed?

Answer 1

$\DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech}$ Try $$ \csch(\theta) = \frac{1}{\sinh(\theta)} = \frac{1/\cosh(\theta)}{\sinh(\theta)/\cosh(\theta)} = \frac{\sech(\theta)}{\tanh(\theta)} = \frac{\sqrt{1 - \tanh^2(\theta)}}{\tanh(\theta)} \, . $$

How did I get this? By cheating, basically. The Wikipedia page for the lemniscate of Gerono has different parametrizations of the curve, one of which is $x = \cos(\varphi), y = \sin(\varphi) \cos(\varphi)$. I used this parametrization to integrate, which gave me an expression involving $\frac{\sqrt{1 - x^2}}{x}$, which led me to this answer.

Answer 2

For positive $x$, another substitution that does not involve integrating $\csc{x}$ or $\operatorname{csch}{x}$ is a hyperbolic secant substitution (similar to a secant/cosecant substitution for $x^2-1$), where $x=\operatorname{sech}{\theta}$, $dx=-\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$, and $$\sqrt{1-x^2}=\sqrt{1-\operatorname{sech}^2{\theta}}=\sqrt{\tanh^2{\theta}}=\tanh{\theta}$$ since $\theta\geq0$ and $\tanh{x}$ is positive for that interval. $$\int{\frac{dx}{x\sqrt{1-x^2}}}=-\int{\frac{\operatorname{sech}{\theta}\tanh{\theta}}{\operatorname{sech}{\theta}\tanh{\theta}}d\theta}=-\int{d\theta}=-\theta+C$$ Substituting $x$ back gives $$-\operatorname{arsech}{x}+C\qquad x>0$$ For negative $x$, the substitution $x=-\operatorname{sech}{\theta}$ and $dx=\operatorname{sech}{\theta}\tanh{\theta}\,d\theta$ give $$-\operatorname{arsech}{\left(-x\right)}+C\qquad x<0$$ Combining the two branches gives $$-\operatorname{arsech}{\left|x\right|}+C$$