Find the arc length (using hyperbolic trig)

In summary, the problem at hand is to find the length of the curve y=ln(x) over the interval 1/2<=x<=2. The attempt at a solution involves using the formula for arc length, but the main issue arises when trying to solve the integral on the right side. The writer tried several methods, including using hyperbolic trig functions and substituting in terms of e, but none of them gave the correct answer. The writer also asks for advice on whether using hyperbolic trig functions is a good approach for this problem.
  • #1
1. The problem statement, all variables and given/known

Find the length of the curve $$y=ln(x),\frac{1}{2}<=x<=2$$

Homework Equations

Using hyperbolic trig isn't necessary, but it's how my text (Serge Lang's A First Course in Calculus) approaches most square roots, and as a result, it's what I've grown accustomed to. I am aware of other solutions (thanks to Wolfram Alpha). I want to know where the heck I'm going wrong.

The Attempt at a Solution

Arc length can be calculated by $$ \int_a^b \sqrt{1+(\frac{dx}{dy})^{2}}dx$$. Since y=ln(x), then dx/dy=1/x dx, which gives our first integral of $$ \int_\frac{1}{2}^2 \sqrt{1+\frac{1}{x}^{2}} dx$$. Combining the two terms under the radical and simplifying, we have $$\int_\frac{1}{2}^2 \frac{\sqrt{x^{2}+1}}{x} dx$$. Next, let x=sinh(t), then dx=cosh(t) dt. Substitution gives us the indefinite integral (with limits of integration to be used after finding an indefinite integral) $$\int \frac {\sqrt{sinh^{2}t+1}}{sinht}dt$$. Renaming sinh^2(t)+1 into cosh^2(t), and then taking the square root of that, then multiplying it by cosht, gives us $$\int \frac {cosh^{2}t}{sinht}dt$$. Renaming cosh^2(t) into sin^2(t)+1, and splitting the fraction into its two equivalent pieces, then simplifying, we now have $$ \int sinh(t)dt+\int \frac{dt}{sinh(t)}$$. The left integral simplifies into sinht dt, whose integral is cosht, and this allows us to substitute back for an explicit expression (in terms of x), namely, $$\sqrt{x^{2}+1}$$.

The right side is my issue. I did a number of things, but what worked the best was to rename sinh(t) in terms of e. I simplified the fraction and found that I had $$\int \frac{2 dt}{e^{t}-e^{-t}}$$. I multiplied this integral by the clever one (e^t)/(e^t), which gave me $$2 \int \frac{e^{t}dt}{e^{2t}-1}$$. Because this looked like a very difficult integrand, I substituted back for x, since t=arcsinh(x) (which we know). Then we have $$2 \int \frac {x+\sqrt{x^2+1}dx}{(x+\sqrt{x^2+1})^{2}-1}$$. Simplifying this I found $$2 \int \frac {x+\sqrt{x^2+1}}{2x^{2}+2x\sqrt{x^{2}+1}}$$. Factoring out a 1/2, which cancels the constant on the outside, we have what appears to simplify into a very simple integrand, that is, $$\int \frac{dx}{x}$$.

Evaluating this from 1/2 to 2 would give me the answer $$\frac{\sqrt{5}}{2}-ln(\frac{1}{4})$$. This is approximately equal to 2.504. This is not the right answer. The book's answer is given as $$\frac{\sqrt{5}}{2}+ln(\frac{4+2\sqrt{5}}{1+\sqrt{5}})$$, which is approximately equal to 2.080.

Phew! That was a long write up from me - any help would be appreciated. Also, some advice: Are the hyperbolic trig functions a good or bad idea here? I know we can also use tan^2(theta)+1 to our advantage here as well.
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  • #2
josephgerth said:
Arc length can be calculated by $$ \int_a^b \sqrt{1+(\frac{dx}{dy})^{2}}dx$$. Since y=ln(x), then dx/dy=1/x dx,
you mean dy/dx in both places. And it's just dy/dx = 1/x.
josephgerth said:
dx=cosh(t) dt
Yes, but you left that cosh out of the next line. (It reappears later.)

After solving the indefinite integral, you should check it satisfies the original equation by differentiating back.

Anyway, it's worth knowing how to integrate cosech, sech, sec and cosec.
E.g.for sec,
##\sec(x).dx = \frac{\cos(x).dx}{1-\sin^2(x)} = \frac {ds}{1-s^2}##
Expand using partial fractions.

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