$$\begin{align} L&:=\int_{\theta_1}^{\theta_2}\sqrt{\left({\mathrm{d}x\over\mathrm{d}\theta}\right)^2+\left({\mathrm{d}y\over\mathrm{d}\theta}\right)^2}\,\mathrm{d}\theta \\\\r&:={2\over\theta} \\ &\begin{cases} \theta_1=1/2\\\theta_2=4 \end{cases} \\&\begin{cases} x=r\cos(\theta)\\y=r\sin(\theta) \end{cases} \\ L&=\int_{{1\over 2}}^{4}\sqrt{\left({\mathrm{d}x\over\mathrm{d}\theta}\right)^2+\left({\mathrm{d}y\over\mathrm{d}\theta}\right)^2}\,\mathrm{d}\theta \end{align}$$

The problem above is from this book(A First Course in Calculus by Serge Lang)

The book says use $~\theta=\sinh(t)~$as a hint.

I tried to find out$~L~$without using that substitution of variable, but failed to.

So I am now willing to use$~\theta=\sinh(t)~$

$$\begin{align} t&:=\sinh^{-1}(\theta)\implies \theta=\sinh(t)\\ {\mathrm{d} \theta \over \mathrm{d} t }&=\cosh(t) \\x&= {2 \over \theta }\cos(\theta)\\ y&= {2 \over \theta }\sin(\theta) \end{align}$$

Stucked from here.

How can I handle trigonometric functions with hyperbolic functions in this case?

  • $\begingroup$ I think I can proceed without substituting variable. Just separating into 2 integrals by ranges of $~\theta~$ may work. $\endgroup$ Jun 2, 2022 at 8:42

1 Answer 1


Substituting the formulas for $x, y$ in the arc length formula and simplifying gives $$L = 2 \int_\frac{1}{2}^4 \frac{\sqrt{1 + \theta^2}}{\theta^2} \,d\theta ;$$ notice that trigonometric functions do not appear in the integrand. (Remark It's easier to arrive at this formula using the equation for arc length of a polar curve $r(\theta)$, namely, $\int_a^b \sqrt{r^2 + \left(\frac{dr}{d\theta}\right)^2}\,d\theta$.)

At this point, the appearance of the factor $\sqrt{1 + \theta^2}$ suggests the substitution $$\theta = \sinh t, \qquad d\theta = \cosh t\,dt ,$$ since $\sqrt{(\sinh t)^2 + 1} = \cosh t$, which transforms the integral to $$2 \int_{\operatorname{arsinh} \frac{1}{2}}^{\operatorname{arsinh} 4} \coth^2 t \,dt = t - \coth t \Bigg\vert_{\operatorname{arsinh} \frac{1}{2}}^{\operatorname{arsinh} 4}.$$ To simplify the resulting expression it will be helpful to recall the identity $\coth \operatorname{arsinh} \theta = \frac{\sqrt{1 + \theta^2}}{\theta}$.

Remark You might find the original integral easier to evaluating using the similar substitution $\theta = \tan \alpha$, $d\theta = \sec^2 \alpha \,d\alpha$, which yields $$2 \int_{\arctan \frac{1}{2}}^{\arctan 4} \frac{d\alpha}{\sin^2 \alpha \cos \alpha} .$$

  • $\begingroup$ I've reached till before $\coth\left(\operatorname{arcsinh}(\theta)\right)=\frac{\sqrt{1+\theta^2}}{\theta}~$. I've been trying to derive it. $\endgroup$ Jun 3, 2022 at 1:28
  • $\begingroup$ One option is to write out both expressions in terms of logarithmic and exponential functions, namely, $\coth t = \frac{e^t + e^{-t}}{e^t - e^{-t}}$, $\operatorname{arsinh}(\theta) = \log\left(\theta + \sqrt{1 + \theta^2}\right)$, substitute, and simplify. It may be helpful to rationalize $\exp (-\operatorname{arsinh} t) = \frac{1}{\theta + \sqrt{1 + \theta^2}}$. $\endgroup$ Jun 3, 2022 at 7:32

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