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I have been struggling with an arc length question, and I want to make sure I get this right. I have the function of:

\begin{align} f(x) = \sqrt{7.2 (x-\frac {1}{7}}) - 2.023, [0.213, 0.127]. \end{align}

I have found the derivative of the function and set up my integral this way:

\begin{align} I &= \int_{0.127}^{0.213} \sqrt{1 + \frac{12.96}{7.2x-\frac{7.2}{7}}}~dx \end{align}

Letting A = 12.96 and simplifying:

\begin{align} I &= \int_{0.127}^{0.213} \sqrt{\frac{7.2x-\frac{7.2}{7}+A}{7.2x-\frac{7.2}{7}}}~dx \end{align}

$u=7.2x-\frac{7.2}{7}, du= 7dx, dx=\frac{du}{7}$:

\begin{align} I &= \int_a^b \sqrt{\frac{{u}+A}{u}}~\frac{du}{7} \end{align}

\begin{align} I &= \frac{1}{7}\int_a^b \sqrt{\frac{{u}+A}{u}}~du \end{align}

$u = C\tan^2v\\ du = 2C \tan v \sec^2 v ~ dv$

\begin{align} I &= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{u + A}{u}}~du\\ &= \frac{1}{7}\int_{x=a}^{x=b} \sqrt{\frac{A(\tan^2 v + 1)}{A \tan^2 v}}~2A\tan v \sec^2 v ~ dv\\ &= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\sec^2 v}{\tan^2 v}}\tan v \sec^2 v ~ dv; & \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{\frac{1}{\cos^2 v}}{\frac{\sin^2 v}{\cos^2 v}}}\frac{\sin v}{\cos^3 v}~ dv; \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \sqrt{\frac{1}{\sin^2 v}}\frac{\sin v}{\cos^3 v}~ dv \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\sin v}\frac{\sin v}{\cos^3 v}~ dv \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{\cos^4 v}~ dv \\ &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-\sin^2(v))^2} dv~ \end{align}

This is where I am stuck. Could I make a substitution such as:

$t = \sin v\\dt=\cos v\ dt\\\frac{dt}{cos\ v}=dv$

and then:

\begin{align} &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{\cos v}{(1-t^2)^2} \frac{dt}{cos\ v}~\\ &= \frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{(1-t^2)^2} dt~ \end{align}

which gives me an ordinary partial fractions integral.

Could I make this substitution or is it not possible because I would have 2 different variables in my integral, and if it's not possible, how else could I solve this integral?

Coderless
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2 Answers2

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I amssuming you work prior to the $u$-substitution is correct.

Instead of using the substitution $u=C\tan^2(\nu)$, consider the more simple one $u=\nu^2$, which reduces your integral to $$I=\frac{2\sqrt{A}}7\int_{\sqrt{a}}^{\sqrt{b}}\sqrt{1+\left(\frac{\nu}{\sqrt{A}}\right)^2}d\nu.$$ There are various approaches of evaluating the integral of $\sqrt{1+x^2}$. One way is to let $\nu=\sqrt{A}\text{ sinh}(s)$.

kiyomi
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  • I can see how that would make my life easier. Does my solution work though, or can I not have two different variables in the same integral? – Coderless Oct 21 '20 at 01:53
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I actually found the answer to my question.

\begin{align} & I=\frac{2A}{7}\int_{x=a}^{x=b} \frac{1}{\cos^3 v}~ dv \\ & =\frac{2A}{7}\int_{x=a}^{x=b} sec^3v~ dv \\ & =\frac{A}{7}\int_{x=a}^{x=b} \sec v \tan v+ln \lvert\sec v + \tan v\rvert~ dv \\ \end{align}

The new bounds would be:

$arctan\frac{\sqrt{7.2*1.27-\frac{7.2}{7}}}{\sqrt{A}}$ and $arctan\frac{\sqrt{7.2*0.213-\frac{7.2}{7}}}{\sqrt{A}}$

It was much simpler than I thought.

Coderless
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