Chapter 5 - Applications Of The Definite Integral In Geometry, Science, And Engineering - 5.4 Length Of A Plane Curve - Exercises Set 5.4 - Page 376: 28

$$L = \frac{{{\pi ^2}}}{2}$$

$$\eqalign{ & x = \cos t + t\sin t,\,\,\,\,\,y = \sin t - t\cos t,\,\,\,\,\left( {0 \leqslant t \leqslant \pi } \right) \cr & {\text{Calculate the derivatives }}\frac{{dx}}{{dt}}{\text{ and }}\frac{{dy}}{{dt}} \cr & \frac{{dx}}{{dt}} = \frac{d}{{dt}}\left[ {\cos t + t\sin t} \right] = - \sin t + t\cos t + \sin t = t\cos t \cr & \frac{{dy}}{{dt}} = \frac{d}{{dt}}\left[ {\sin t - t\cos t} \right] = \cos t + t\sin t - \cos t = t\sin t \cr & {\text{Use the arc length formula }}L = \int_a^b {\sqrt {{{\left( {\frac{{dx}}{{dt}}} \right)}^2} + {{\left( {\frac{{dy}}{{dt}}} \right)}^2}} dt} \cr & L = \int_0^\pi {\sqrt {{{\left( {t\cos t} \right)}^2} + {{\left( {t\sin t} \right)}^2}} dt} \cr & L = \int_0^\pi {\sqrt {{t^2}{{\cos }^2}t + {t^2}{{\sin }^2}t} dt} \cr & L = \int_0^\pi {\sqrt {{t^2}} dt} \cr & L = \int_0^\pi {tdt} \cr & {\text{Integrating}} \cr & L = \frac{1}{2}\left[ {{t^2}} \right]_0^\pi \cr & L = \frac{1}{2}\left[ {{\pi ^2} - 0} \right] \cr & L = \frac{{{\pi ^2}}}{2} \cr} $$