Answer
$$L = \int_0^\pi {\sqrt {1 + {{\sin }^2}x} dx} $$
Work Step by Step
$$\eqalign{
& y = f\left( x \right) = \int_0^x {\sin t} dt,{\text{ On the interval }}\left[ {0,\pi } \right] \cr
& {\text{Differentiate}} \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {\int_0^x {\sin t} dt} \right] \cr
& {\text{Using the Fundamental Theorem Of Calculus aPart 1}} \cr
& \frac{{dy}}{{dx}} = \sin x \cr
& {\text{Use the formula }}L = \int_c^d {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} d} x \cr
& L = \int_0^\pi {\sqrt {1 + {{\left( {\sin x} \right)}^2}} dx} \cr
& {\text{Simplifying}} \cr
& L = \int_0^\pi {\sqrt {1 + {{\sin }^2}x} dx} \cr} $$
