Area Solution 9

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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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