4
$\begingroup$

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?

$\endgroup$
  • 2
    $\begingroup$ I think $x = \sin t$ seems like a better substitute. $\endgroup$ – Jacky Chong Jan 8 '17 at 3:42
  • 2
    $\begingroup$ 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. $\endgroup$ – André 3000 Jan 8 '17 at 3:56
  • $\begingroup$ sinh or cosh of t seems better to substitute $\endgroup$ – Saketh Malyala Jan 8 '17 at 4:31
  • $\begingroup$ @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. $\endgroup$ – W. Zhu Jan 8 '17 at 4:38
3
$\begingroup$

$\DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\sech}{sech}$ 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.

$\endgroup$
  • $\begingroup$ Nice trick. I read the page but do not know how to integrate using parametrization. $\endgroup$ – W. Zhu Jan 8 '17 at 4:27
  • $\begingroup$ @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. $\endgroup$ – André 3000 Jan 8 '17 at 5:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.