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

2 Answers
Jun 23, 2018

2*arc coth(sqrt(3))

Explanation:

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)

Jun 25, 2018

L=ln(2+sqrt3) units.

Explanation:

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)