:browse confirm e Area Solution 9

tex2html_wrap_inline125



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

Click HERE to return to the list of problems.