How to calculate the arc length of a difficult radical function 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?

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.
