Answer:
Length of curve, L=0.3828
Step-by-step explanation:
We are given a curve [tex]y=\ln \cos(x)[/tex]
Length of arc L=[tex]\int ds[/tex]
Where,
[tex]ds=\sqrt{1+y'^2}dx[/tex]
[tex]\text{Where, }0\leq x \leq \frac{\pi}{4}[/tex]
[tex]y=\ln \cos(x)[/tex]
Derivative of y
[tex]y'=-\tan x[/tex]
Substitute y' into ds
[tex]ds=\sqrt{1+\tan^2x}dx\Rightarrow \int |\sec x|dx[/tex]
[tex]\text{Length of arc L}=\int ds[/tex]
[tex]L=\int_{0}^{\pi/4} |\sec x|dx[/tex]
[tex]L=\log|\tan x+\sec x|_{0}^{\pi/4} [/tex]
[tex]L=\log|1+\sqrt{2}|\approx 0.3828[/tex]
Thus, Length of curve, L=0.3828
