# How do I find the arc length of the curve y=ln(cos(x)) over the interval [0,π/4]?

Feb 1, 2015

The answer is: $\ln \left(\sqrt{2} + 1\right)$

To find the lenght of a curve $L$, written in cartesian coordinates, it is necessary to use this formula:

$L = {\int}_{a}^{b} \sqrt{\left(1 + {\left[f ' \left(x\right)\right]}^{2}\right)} \mathrm{dx}$.

Since $f ' \left(x\right) = \frac{1}{\cos} x \cdot \left(- \sin x\right)$, then:

$L = {\int}_{0}^{\frac{\pi}{4}} \sqrt{1 + \frac{{\sin}^{2} x}{{\cos}^{2} x}} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \sqrt{\frac{{\cos}^{2} x + {\sin}^{2} x}{{\cos}^{2} x}} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \sqrt{\frac{1}{{\cos}^{2} x}} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \frac{1}{\cos} x \mathrm{dx}$.

This integral has to be done using this substitution (parametric formulae):

$t = \tan \left(\frac{x}{2}\right) \Rightarrow \frac{x}{2} = \arctan t \Rightarrow x = 2 \arctan x \Rightarrow \mathrm{dx} = \frac{2}{1 + {t}^{2}} \mathrm{dt}$,

and it's known that: $\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$,

if $x = 0$ then $t = 0$

if $x = \frac{\pi}{4}$ then $t = \tan \left(\frac{\pi}{8}\right) = \sqrt{2} - 1$

So:

${\int}_{0}^{\sqrt{2} - 1} \frac{1}{\cos} x \mathrm{dx} = {\int}_{0}^{\sqrt{2} - 1} \frac{1}{\frac{1 - {t}^{2}}{1 + {t}^{2}}} \frac{2}{1 + {t}^{2}} \mathrm{dt} = 2 {\int}_{0}^{\sqrt{2} - 1} \frac{1}{1 - {t}^{2}} \mathrm{dt}$,

$\frac{1}{1 - {t}^{2}} = \frac{1}{\left(1 + t\right) \left(1 - t\right)} = \frac{A}{1 + t} + \frac{B}{1 - t} = \frac{A \left(1 - t\right) + B \left(1 + t\right)}{\left(1 + t\right) \left(1 - t\right)}$,

Two polynomials ($1$ on the left and $\left[A \left(1 - t\right) + B \left(1 + t\right)\right]$ on the right) are identical if they assume the same values at the same values of $t$:

If $t = - 1$ then $1 = A \cdot 2 \Rightarrow A = \frac{1}{2}$;

If $t = 1$ then $1 = B \cdot 2 \Rightarrow B = \frac{1}{2}$.

The integral becomes:

$2 {\int}_{0}^{\sqrt{2} - 1} \left[\frac{\frac{1}{2}}{1 + t} + \frac{\frac{1}{2}}{1 - t}\right] \mathrm{dt} = 2 \left(\frac{1}{2}\right) {\int}_{0}^{\sqrt{2} - 1} \left[\frac{1}{1 + t} - \frac{- 1}{1 - t}\right] \mathrm{dt} = {\left[\ln | 1 + t | - \ln | 1 - t |\right]}_{0}^{\sqrt{2} - 1} = \ln | 1 + \sqrt{2} - 1 | - \ln | 1 - \sqrt{2} + 1 | = \ln \sqrt{2} - \ln \left(2 - \sqrt{2}\right) = \ln \left(\frac{\sqrt{2}}{2 - \sqrt{2}}\right) = \ln \left(\frac{\sqrt{2}}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}}\right) = \ln \left(\frac{2 \sqrt{2} + 2}{4 - 2}\right) = \ln \left(\frac{2 \left(\sqrt{2} + 1\right)}{2}\right) = \ln \left(\sqrt{2} + 1\right)$.