:browse confirm e Area Solution 9

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 $0 \le x \le \pi/3$.) $$= \displaystyle{ \ln \Big| \sec x + \tan x \Big| \ \Bigg\vert_{0}^{\pi/3} }$$ $$= \displaystyle{ \ln \Big| \sec (\pi/3) + \tan (\pi/3) \Big| - \ln \Big| \sec 0 + \tan 0 \Big| }$$ $$= \displaystyle{ \ln \Big| 2 + \sqrt{3} \Big| - \ln \Big| 1 + 0 \Big| }$$ $$= \ln(2 + \sqrt{3}) - 0$$ $$= \ln(2 + \sqrt{3})$$