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

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?

• I think $x = \sin t$ seems like a better substitute. Commented Jan 8, 2017 at 3:42
• Just as a remark, the denominator of the integrand defines a plane curve called the lemniscate of Gerono. If you weren't required to make a hyperbolic substitution, I would have recommended one using the rational parametrization given on the linked page. Commented Jan 8, 2017 at 3:56
• sinh or cosh of t seems better to substitute Commented Jan 8, 2017 at 4:31
• @JackyChong With $x=\sin t$, the integral becomes a similar expression $-\ln|\csc t+\cot t|+C$. We have a similar problem, too, if we use a smiliar approach, i.e., $\cot^2 t=\csc^2 t-1$. We need a workaround to reach $\cot t=\frac{\sqrt{1-x^2}}{x}$. So, I think it is just a matter of preference. We are just more comfortable with trigonometric functions. Commented Jan 8, 2017 at 4:38

$\DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech}$ 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.
• @W.Zhu You can interpret the given integral as the path integral $\int_C \frac{dx}{y}$ over the curve $C$ in the plane given by $y = x \sqrt{1-x^2}$ or better yet, $y^2 = x^2(1-x^2)$. Then you substitute in the given parametrizations $x = \cos(\varphi), y = \sin(\varphi) \cos(\varphi)$. In the end, it's equivalent to the substituting $x = \cos(\varphi)$ into the original integral. Commented Jan 8, 2017 at 5:08
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 $$-\operatorname{arsech}{\left|x\right|}+C$$