Answer
$$ \approx 3.82019$$
Work Step by Step
$$\eqalign{
& y' = {\sin ^2}x,{\text{ }}\left[ {0,\pi } \right] \cr
& {\text{Use the length arc formula }} \cr
& L = \int_a^b {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} dx} {\text{ }}\left( {\bf{1}} \right) \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\sin }^2}x} \right] = 2\sin x\cos x,{\text{ }}\left[ {0,\pi } \right] \to a = 0,{\text{ }}b = \pi \cr
& \frac{{dy}}{{dx}} = \sin 2x \cr
& {\text{Substitute }}\frac{{dy}}{{dx}}{\text{ into }}\left( {\bf{1}} \right) \cr
& L = \int_0^\pi {\sqrt {1 + {{\left( {\sin 2x} \right)}^2}} dx} \cr
& L = \int_0^\pi {\sqrt {1 + {{\sin }^2}2x} dx} \cr
& {\text{Approximate by using a CAS}} \cr
& L = \int_0^\pi {\sqrt {1 + {{\sin }^2}2x} dx} \approx 3.82019 \cr} $$
