$\underset{0}{\overset{\pi }{\int }}|\cos \theta -\sin \theta |d\theta =$

$\underset{0}{\overset{\pi }{\int }}|\cos \theta -\sin \theta |d\theta$

$\left(\begin{array}{l}\because Make \cos \theta -\sin \theta =0\\ \Rightarrow \cos \theta =\sin \theta \\ \Rightarrow \theta =\frac{\pi }{4}\end{array}\right)$

$\therefore I=\underset{0}{\overset{\pi /4}{\int }}|\cos \theta -\sin \theta |+\underset{\pi /4}{\overset{\pi }{\int }}|\cos \theta -\sin \theta |d\theta$

$=\underset{0}{\overset{\pi /4}{\int }}(\cos \theta -\sin \theta )d\theta +\underset{\pi /4}{\overset{\pi }{\int }}-(\cos \theta -\sin \theta )d\theta$

$\left(\because \cos \theta >\sin \theta if \theta \in \left(0,\frac{\pi }{4}\right)\right)$

$\left(\because \cos \theta <\sin \theta if \theta \in \left(\frac{\pi }{4},\pi \right)\right)$

$={(\sin \theta +\cos \theta )}_0^{\pi /4}+{(-\sin \theta -\cos \theta )}_{\pi /4}^{\pi }$

$=\left(\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}+\frac{{\displaystyle 1}}{{\displaystyle \sqrt{2}}}\right)-(0+1)+\left(\sin \frac{{\displaystyle \pi }}{{\displaystyle 4}}+\cos \frac{{\displaystyle \pi }}{{\displaystyle 4}}\right)-(\sin \pi +\cos \pi )$

$=\left(\sqrt{2}-1)+(\sqrt{2}-(0-1)\right)$

$=2\sqrt{2}$