Answer
$\ln(\sqrt 2+1)$
Work Step by Step
As we are given that $r(t)=\cos ti+\sin tj+ln(\cos t)k$ ; $0 \leq t \leq \frac{\pi}{4}$
Length of the curve can be obtained by using formula, such as $L=\int_a^b |r'(t)| dt$
Now,$r'(t)=\lt -\sin t, \cos t, \dfrac{1}{\cos t}(-\sin t)\gt=\lt -\sin t, \cos t, -\tan t\gt$; $|r'(t)|=\sqrt {(\sin t)^2+( \cos t)^2+(-\tan t)^2}dt$
or, $=\sqrt{ 1+\tan^2 t}$
or, $=\sec t$
Since, $L=\int_{0}^\frac{\pi}{4} \sec tdt=ln(\sec t+ \tan t)_{0}^\frac{\pi}{4}$
or, $=ln(\sec(\frac{\pi}{4})+ \tan (\frac{\pi}{4}))-ln(\sec (0)+ \tan (0))$
This yields,
$L=\ln(\sqrt 2+1)$
