Find the arc length of the function below?

y=\ln(\secx), with parameters 0\lex\le\pi/4?

1 Answer
Apr 24, 2018

The arc has a length of ln(sqrt2+1).

Explanation:

For a function in the form y=f(x), the arc length from [a,b] is given by

L=int_a^bsqrt(1+(dy/dx)^2)dx.

The given function y=lnsecx is in the form y=f(x), so we'll use the above formula. Furthermore, [a,b]=[0,pi/4], so these are our integral's bounds.

Differentiate using the Chain Rule:

y=lnsecx
(dy)/dx=1/secx*d/dxsecx

dy/dx=(cancelsecxtanx)/(cancelsecx)

dy/dx=tanx

Then,

L=int_0^(pi/4)sqrt(1+tan^2x)dx

Recall the identity 1+tan^2x=sec^2x. Then we get

L=int_0^(pi/4)sqrt(sec^2x)dx

L=int_0^(pi/4)secxdx

This is a common integral and worth memorizing. In general,

intsecxdx=ln|secx+tanx|+C

Then,

int_0^(pi/4)secxdx=ln(secx+tanx)|_0^(pi/4)

Absolute value bars are dropped because secant and tangent are positive on [0, pi/4].

=ln(sec(pi/4)+tan(pi/4))-ln(sec(0)+tan(0))

=ln(sqrt2+1)-ln(1)

=ln(sqrt2+1)

The arc has a length of ln(sqrt2+1).