Answer
$$L = \sqrt 2 \pi $$
Work Step by Step
$$\eqalign{
& x = \sin t + \cos t,{\text{ }}y = \sin t - \cos t{\text{ on the interval }}\left( {0 \leqslant t \leqslant \pi } \right) \cr
& {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr
& \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\sin t + \cos t} \right] = \cos t - \sin t \cr
& \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\sin t - \cos t} \right] = \cos t + \sin t \cr
& {\text{Use the Arc Length Formula For Parametric Curves}} \cr
& L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr
& L = \int_0^\pi {\sqrt {{{\left( {\cos t - \sin t} \right)}^2} + {{\left( {\cos t + \sin t} \right)}^2}} dt} \cr
& L = \int_0^\pi {\sqrt {1 - 2\sin t\cos t + 1 + 2\sin t\cos t} dt} \cr
& L = \int_0^\pi {\sqrt 2 dt} \cr
& L = \int_0^\pi {\sqrt 2 dt} \cr
& {\text{Integrating}} \cr
& L = \sqrt 2 \left[ t \right]_0^\pi \cr
& L = \sqrt 2 \left( {\pi - 0} \right) \cr
& L = \sqrt 2 \pi \cr} $$
