Answer
ln(1+$\sqrt 2$)
Work Step by Step
The arc length L=$\int$ √[1+$(dy/dx)^{2}$] dx
y=ln(cosx) dy/dx=$\frac{-sinx}{cosx}$=-tanx
L=$\int$ √[1+$(-tanx)^{2}$] dx = $\int$ √($sec^{2}x$) dx=$\int$secx dx (secx>0 when x is between 0 and $\frac{\pi}{4}$)
Now $\int$secx dx=ln|tanx + secx|
When x∈[0, $\frac{\pi}{4}$], L=ln|tan$\frac{\pi}{4}$ + sec$\frac{\pi}{4}$| -ln|tan0+sec0|=ln(1+$\sqrt 2$)
