SOLUTION 9: $ \ \ $ If $ y = \displaystyle{ \ln(\cos x) } $ for $ 0 \le x \le \pi/3 $, then
$$ \displaystyle{ { dy \over dx} = { 1 \over \cos x} (-\sin x) = { - \sin x \over \cos x } = -\tan x } $$
so that
$$ ARC = \displaystyle{ \int_{0}^{\pi/3} \sqrt{ 1 + \Big({dy \over dx}\Big)^2 } \ dx } $$
$$ = \displaystyle{ \int_{0}^{\pi/3} \sqrt{ 1 + ( - \tan x )^2 } \ dx } $$
$$ = \displaystyle{ \int_{0}^{\pi/3} \sqrt{ 1 + \tan^2 x } \ dx } $$
$$ = \displaystyle{ \int_{0}^{\pi/3} \sqrt{ \sec^2x } \ dx } $$
(Recall that $ \sqrt{z^2} = |z| $.)
$$ = \displaystyle{ \int_{0}^{\pi/3} \Big| \sec x \Big| \ dx } $$
(The $ \sec x > 0$ for $ 0 \le x \le \pi/3 $.)
$$ = \displaystyle{ \int_{0}^{\pi/3} \sec x \ dx } $$
$$ = \displaystyle{ \ln \Big| \sec x + \tan x \Big| \ \Bigg\vert_{0}^{\pi/3} } $$
$$ = \displaystyle{ \ln \Big| \sec (\pi/3) + \tan (\pi/3) \Big| - \ln \Big| \sec 0 + \tan 0 \Big| } $$
$$ = \displaystyle{ \ln \Big| 2 + \sqrt{3} \Big| - \ln \Big| 1 + 0 \Big| } $$
$$ = \ln(2 + \sqrt{3}) - 0 $$
$$ = \ln(2 + \sqrt{3}) $$
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