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

Feb 25, 2015

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

The lenght of a function written in cartesian coordinates is:

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

So:

$y ' = \frac{1}{\cos} x \cdot \left(- \sin x\right)$

Than:

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

$= {\int}_{0}^{\frac{\pi}{4}} \sqrt{\frac{1}{\cos} ^ 2 x} \mathrm{dx} = {\int}_{0}^{\frac{\pi}{4}} \frac{1}{\cos} x \mathrm{dx} = \left(1\right)$

Now we have to make a substituition (with tangent half-angle 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

if $x = 0$ than $t = \tan 0 = 0$

if $x = \frac{\pi}{4}$ than, for the half angle-formula,
$t = \tan \left(\frac{\pi}{8}\right) = \sin \frac{\frac{\pi}{4}}{1 + \cos \left(\frac{\pi}{4}\right)} = \frac{\frac{\sqrt{2}}{2}}{1 + \frac{\sqrt{2}}{2}} =$

$= \frac{\frac{\sqrt{2}}{2}}{\frac{2 + \sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \cdot \frac{2}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2 + \sqrt{2}} =$

$\frac{\sqrt{2}}{2 + \sqrt{2}} \cdot \frac{2 - \sqrt{2}}{2 - \sqrt{2}} = \frac{2 \sqrt{2} - 2}{4 - 2} = \frac{2 \left(\sqrt{2} - 1\right)}{2} =$

$= \sqrt{2} - 1$

So, remembering that $\cos x = \frac{1 - {t}^{2}}{1 + {t}^{2}}$:

$\left(1\right) = {\int}_{0}^{\sqrt{2} - 1} \frac{1}{\frac{1 - {t}^{2}}{1 + {t}^{2}}} \cdot \frac{2}{1 + {t}^{2}} \mathrm{dt} =$

$= 2 {\int}_{0}^{\sqrt{2} - 1} \frac{1}{1 - {t}^{2}} \mathrm{dt} = \left(2\right)$

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

and 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 $1 = 2 A \Rightarrow A = \frac{1}{2}$;

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

So:

$\left(2\right) = 2 {\int}_{0}^{\sqrt{2} - 1} \left(\frac{\frac{1}{2}}{1 - t} + \frac{\frac{1}{2}}{1 + t}\right) \mathrm{dt} =$

$= 2 \cdot \frac{1}{2} {\int}_{0}^{\sqrt{2} - 1} \left(\frac{1}{1 - t} + \frac{1}{1 + t}\right) \mathrm{dt} =$

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

=[-ln|1-(sqrt2-1)|+ln|1+(sqrt2-1)|-(ln|1-0|+ln|1+0|]=

$= - \ln \left(2 - \sqrt{2}\right) + \ln \left(\sqrt{2}\right) = \ln \left(\frac{\sqrt{2}}{2 - \sqrt{2}}\right) =$

$= \ln \left(\frac{\sqrt{2}}{2 - \sqrt{2}} \cdot \frac{2 + \sqrt{2}}{2 + \sqrt{2}}\right) =$

$= \ln \left(\frac{2 \sqrt{2} + 2}{4 - 2}\right) = \ln \left(\frac{2 \left(\sqrt{2} + 1\right)}{2}\right) = \ln \left(\sqrt{2} + 1\right)$.