Вычислить длину дуги, заданной параметрически. L=

$\int\limits^\pi_0\ {sqrt((x'(t))^2+(y'(t))^2)} \, dt =$

= $\int\limits^\pi_0\ {sqrt((2(t-sint))')^2+(2(1-cos t))')^2)} \, dt =$

= $\int\limits^\pi_0\ {sqrt((2-2cost)^2+(2sin t)^2)} \, dt =$

= $\int\limits^\pi_0\ {sqrt(4-8cos t+4cos^2 t+4sin^2 t)} \, dt =$

= $\int\limits^\pi_0\ {sqrt(4-8cos t+4)} \, dt =$

= $\int\limits^\pi_0\ {sqrt(8-8cost)} \, dt =$

= $sqrt(8) \int\limits^\pi_0\ {sqrt(1-cos t)} \, dt =$

= $sqrt(8)\int\limits^\pi_0\ {sqrt(2sin^2 (t/2))} \, dt =$

= $sqrt(8)*2*sqrt(2) \int\limits^\pi_0\ {sin (t/2)} \, dt/2 =$

= $8 *(-cos (t/2)) |\limits^\pi_0\ =$

= $8 (-cos (pi/2)+cos (0/2))=8*(-0+1)=8$