Срочно Решите пожалуйста Определенный интеграл [tex] \int\limits^2_1 { \sqrt{4-x^2} } \, dx [/tex]. Created by Franklin90. matematika-ru

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Срочно Решите пожалуйста
Определенный интеграл [tex] \int\limits^2_1 { \sqrt{4-x^2} } \, dx [/tex]

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

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

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