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

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Answer 1 · 2018-06-23T09:18:17

#2*arc coth(sqrt(3))#

#f'(x)=-tan(x)#
so we have

#int_0^(pi/3)sqrt(1+tan^2(x))dx=2arc coth(sqrt(3))#
Note that

#1+tan^2(x)=(sin^2(x)+cos^2(x))/cos^2(x)=1/cos^2(x)#

Answer 2 · 2018-06-25T04:02:04

#L=ln(2+sqrt3)# units.

#y=ln(cosx)#

#y'=-tanx#

Arc length is given by:

#L=int_0^(pi/3)sqrt(1+tan^2x)dx#

Simplify:

#L=int_0^(pi/3)secxdx#

Integrate directly:

#L=[ln|secx+tanx|]_0^(pi/3)#

Insert the limits of integration:

#L=ln(2+sqrt3)#