# How do you find the arc length of the curve y=lncosx over the interval [0, pi/3]?

Jun 23, 2018

$2 \cdot a r c \coth \left(\sqrt{3}\right)$

#### Explanation:

$f ' \left(x\right) = - \tan \left(x\right)$
so we have

${\int}_{0}^{\frac{\pi}{3}} \sqrt{1 + {\tan}^{2} \left(x\right)} \mathrm{dx} = 2 a r c \coth \left(\sqrt{3}\right)$
Note that

$1 + {\tan}^{2} \left(x\right) = \frac{{\sin}^{2} \left(x\right) + {\cos}^{2} \left(x\right)}{\cos} ^ 2 \left(x\right) = \frac{1}{\cos} ^ 2 \left(x\right)$

Jun 25, 2018

$L = \ln \left(2 + \sqrt{3}\right)$ units.

#### Explanation:

$y = \ln \left(\cos x\right)$

$y ' = - \tan x$

Arc length is given by:

$L = {\int}_{0}^{\frac{\pi}{3}} \sqrt{1 + {\tan}^{2} x} \mathrm{dx}$

Simplify:

$L = {\int}_{0}^{\frac{\pi}{3}} \sec x \mathrm{dx}$

Integrate directly:

$L = {\left[\ln | \sec x + \tan x |\right]}_{0}^{\frac{\pi}{3}}$

Insert the limits of integration:

$L = \ln \left(2 + \sqrt{3}\right)$