# Arc Length integration always unusually difficult? Any special trick?

1. Oct 30, 2012

### B3NR4Y

I don't understand why finding the arc length is always difficult. I understand the formula and know pretty much all the integration methods, but whenever I try to find the arc length of a function like 8x2 = 27y3 from 1 to 8 it's unusually difficult.
I would start by solving for x

\begin{split}
&\frac{2}{3} x^{\frac{2}{3}} = y\\
&\frac{4}{9x^{\frac{1}{3}}} = y'\\
\end{split}

Now I add this into the integration formula $\int_a^b \sqrt{1+(y')^{2}} \, dx$
$$\begin{split} L^{8}_{1} &= \int_1^8 \, \sqrt{1+ (\frac{4}{9x^{\frac{1}{3}}})^{2}} \, dx \\ &= \int_1^8 \, \sqrt{1+ (\frac{16}{81x^{\frac{2}{3}}})} \, dx \\ &= \int_1^8 \sqrt{\frac{81x^{\frac{2}{3}}+16}{81x^{\frac{2}{3}}}} \, dx \\ &= \int_1^8 \frac{\sqrt{81x^{\frac{2}{3}}+16}}{9x^{\frac{1}{3}}} \, dx \end{split}$$
From here I would do a u-substitution of $u=81x^{\frac{2}{3}}+16$ and then take the derivative to find $\frac{1}{54} du = \frac{1}{x^{\frac{1}{3}}} dx$
My new limits would be 97 and 340

\begin{split}
L^{8}_{1} &= \frac{1}{9*54} \int_{97}^{340} \sqrt{u} \, du \\
&= \frac{1}{486} \Bigg(\frac{3}{2} u^{\frac{3}{2}} |^{97}_{340} \Bigg) \\
&=7.28... \\
\end{split}

I got the question right, by going slowly and typing it all out, derp. Anyway, is there a special trick for these types of problems? Or is it to just go slow and remember rules.

Last edited by a moderator: Oct 30, 2012
2. Oct 31, 2012

### lurflurf

Those are often messy because of the form of the integrals.
Many calculus books choose y to make the integral simple.
Even simple y live x^3, sin(x), and 1/x give non-elementary integrals.

3. Oct 31, 2012

### arildno

Finding the arc length of the hyperbolic cosine curve is, however, fairly trivial.