Решить определенный интеграл: [tex] \int\limits^3_0 { x^2 \sqrt{9-x^2} } \, dx [/tex]. Created by Cate1. matematika-ru

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Решить определенный интеграл: [tex] \int\limits^3_0 { x^2 \sqrt{9-x^2} } \, dx [/tex]

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$\int _0^3\, x^2\sqrt{9-x^2}dx=[\, x=3sint,\, dx=3cost\, dt,\,\\\\ 9-x^2=9-9sin^2t=9(1-sin^2t)=9cos^2t,\; sint=\frac{x}{3},\; t=arcsin\frac{x}{3}\\\\x_1=0\; \to \; t_1=arcsin0=0,\; x_2=3 \; \to \; t_2=arcsin1=\frac{\pi}{2} ]=\\\\=\int _0^{\frac{\pi}{2}}9sin^2t\sqrt{9cos^2t}\cdot 3cost\, dt=\int _0^{\frac{\pi}{2}}\, 81sin^2t\cdot cos^2t\, dt=\\\\=81\int _0^{\frac{\pi}{2}}(sint\cdot cost)^2dt=[\, sint\cdot cost=\frac{1}{2}sin2t\, ]=81\int _0^{\frac{\pi}{2}}\frac{1}{4}sin^22t\, dt=$

$=\frac{81}{4}\int _0^{\frac{\pi}{2}}\, \frac{1-cos4t}{2}dt=\frac{81}{8}(t-\frac{1}{4}sin4t)|_0^{\frac{\pi}{2}}=\frac{81}{8}(\frac{\pi}{2}-0)=\frac{81}{16}\pi$

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