# What is the arc length of f(x)=secx*tanx  in the interval [0,pi/4]?

Aug 4, 2016

The arc length formula is basically the distance formula adapted to a function. Whereas the distance formula would be $D = \sqrt{{\left(\Delta x\right)}^{2} + {\left(\Delta y\right)}^{2}}$, the arc length formula is:

$\setminus m a t h b f \left(s = {\int}_{a}^{b} \sqrt{1 + {\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}} \mathrm{dx}\right)$

So, we need to take the derivative and then square it.

$\frac{d}{\mathrm{dx}} \left[\sec x \tan x\right]$

$= \sec x \cdot {\sec}^{2} x + \sec x {\tan}^{2} x$

$= \sec x \left({\sec}^{2} x + {\tan}^{2} x\right)$

$= \sec x \left(1 + 2 {\tan}^{2} x\right)$

And squaring it...

${\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)}^{2}$

$= \textcolor{g r e e n}{{\sec}^{2} x {\left(2 {\tan}^{2} x + 1\right)}^{2}}$

And now putting it in the integral.

$s = \textcolor{b l u e}{{\int}_{0}^{\pi \text{/} 4} \sqrt{1 + {\sec}^{2} x {\left(2 {\tan}^{2} x + 1\right)}^{2}} \mathrm{dx}}$

At that point, you should evaluate the integral numerically on your calculator. A TI-89 should be able to do it, but working it out would be far too complicated.

I get $\approx \textcolor{b l u e}{1.6458}$ using Wolfram Alpha, and this integral returns the same answer.

As you can see, the setup is usually pretty easy. What's usually hard (and generally nearly impossible on medium-complexity functions) is the actual simplification work.