Question

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Определенный интеграл $intlimits^2_1 { sqrt{4-x^2} } , dx$

Answer 1

$intlimits^2_1 {sqrt{4-x^2}} , dx =[; x=2sint, ; dx=2cost, dt,; t=arcsinfrac{x}{2},\\t_1=arcsinfrac{1}{2}=frac{pi}{6},; t_2=arcsin1=frac{pi}{2}, ]=\\intlimits _{frac{pi}{6}}^{frac{pi}{2}}; sqrt{4-4sin^2t}cdot 2cost, dt= 2cdot intlimits^{frac{pi}{2}}_{frac{pi}{6}} sqrt{4cos^2t}cdot cost , dt =\\=4cdot intlimits^{frac{pi}{2}}_{frac{pi}{6}} {cos^2t} , dt =4cdot intlimits^{frac{pi}{2}}_{frac{pi}{6}} { frac{1+cos2t}{2} } , dx =$

$=2cdot intlimits^{frac{pi}{2}}_{frac{pi}{6}} {(1+cos2t)} , dt =2cdot (t+frac{1}{2}sin2t)|limits _{frac{pi}{6}}^{frac{pi}{2}}=\\=2cdot (frac{pi}{2}+frac{1}{2}sinpi )-2cdot (frac{pi}{6}+frac{1}{2}sinfrac{pi}{3})=pi -frac{pi}{3}-frac{sqrt3}{2}; .$