Вычислить длину дуги, заданной параметрически.   [latex]\left \{ {{x=2(t-sint)} \atop {y=2(1-cost)}} \right.[/latex]        [latex]0\leq t \leq \pi[/latex]   L=   [latex]\int\limits^\beta_\alpha {\sqrt{[\phi'(t)]^2+[\Psi'(t)]^2...

[latex] \int\limits^\pi_0\ {sqrt((x'(t))^2+(y'(t))^2)} \, dt =[/latex] =[latex]\int\limits^\pi_0\ {sqrt((2(t-sint))')^2+(2(1-cos t))')^2)} \, dt =[/latex] =[latex]\int\limits^\pi_0\ {sqrt((2-2cost)^2+(2sin t)^2)} \, dt =[/latex] =[latex]\int\limits^\pi_0\ {sqrt(4-8cos t+4cos^2 t+4sin^2 t)} \, dt =[/latex] =[latex]\int\limits^\pi_0\ {sqrt(4-8cos t+4)} \, dt =[/latex] =[latex]\int\limits^\pi_0\ {sqrt(8-8cost)} \, dt =[/latex] =[latex]sqrt(8) \int\limits^\pi_0\ {sqrt(1-cos t)} \, dt =[/latex] =[latex]sqrt(8)\int\limits^\pi_0\ {sqrt(2sin^2 (t/2))} \, dt =[/latex] =[latex]sqrt(8)*2*sqrt(2) \int\limits^\pi_0\ {sin (t/2)} \, dt/2 =[/latex] =[latex]8 *(-cos (t/2)) |\limits^\pi_0\ =[/latex] =[latex]8 (-cos (pi/2)+cos (0/2))=8*(-0+1)=8 [/latex]