Ответ:
Пошаговое объяснение:
[tex]\displaystyle\\\int\limits^1_0 {\frac{x^2}{\sqrt{4-x^2} } } \, dx \\[/tex]
Тригонометрическая подстановка:
[tex]\displaystyle\\\frac{x}{2}=sinu\ \ \ \ \Rightarrow\ \ \ \ u=arcsin\frac{x}{2}\\\\x=2*sinu\ \ \ \ (x)^2=(2sinu)^2\ \ \ \ x^2=4sin^2u\ \ \ \ \frac{x^2}{4} =sin^2u\ \ \ \ \frac{x^2}{4}=1-cos^2u\\\\cos^2u=1-\frac{x^2}{4} \ \ \ \ cosu=\sqrt{1-\frac{x^2}{4} } .[/tex]
[tex]\displaystyle\\\int\frac{4*sin^2u*2cosu}{\sqrt{4-4sin^2u} } du=\int\frac{8*sin^2u*cosu}{2*\sqrt{1-sin^2u} }du=\int\frac{4*sin^2u*cosu}{cosu}du=\\\\ =4*\int sin^2u \ du=4*\int \frac{1-cos2u}{2}du=2*\int(1-cos2u)du=2*\int du- 2*\int cos2u \ du=2*u-sin2u=2*u-2*sinu*cosu.[/tex]
Обратная тригонометрическая подстановка:
[tex]\displaystyle\\2*u-2*sinu*cosu=2*arcsin\frac{x}{2} -2*\frac{x}{2} *\sqrt{1-\frac{x^2}{4} }=2*arcsin \frac{x}{2}-x*\sqrt{1-\frac{x^2}{4} } .\\\\\int\limits^1_0 {\frac{x^2}{\sqrt{4-x^2} } } \, dx =2*arcsin\frac{x}{2} \ |^1_0-x*\sqrt{1-\frac{x^2}{4} } \ |^1_0=\\\\=2*arcsin\frac{1}{2} -2*arcsin0-(1*\sqrt{1-\frac{1^2}{4} }-0*\sqrt{1-\frac{0^2}{4} } }) =\\\\=2*\frac{\pi }{6} -0-(\sqrt{\frac{3}{4} } -0)=\frac{\pi }{3} -\frac{\sqrt{3} }{2} .[/tex]
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Answers & Comments
Ответ:
Пошаговое объяснение:
[tex]\displaystyle\\\int\limits^1_0 {\frac{x^2}{\sqrt{4-x^2} } } \, dx \\[/tex]
Тригонометрическая подстановка:
[tex]\displaystyle\\\frac{x}{2}=sinu\ \ \ \ \Rightarrow\ \ \ \ u=arcsin\frac{x}{2}\\\\x=2*sinu\ \ \ \ (x)^2=(2sinu)^2\ \ \ \ x^2=4sin^2u\ \ \ \ \frac{x^2}{4} =sin^2u\ \ \ \ \frac{x^2}{4}=1-cos^2u\\\\cos^2u=1-\frac{x^2}{4} \ \ \ \ cosu=\sqrt{1-\frac{x^2}{4} } .[/tex]
[tex]\displaystyle\\\int\frac{4*sin^2u*2cosu}{\sqrt{4-4sin^2u} } du=\int\frac{8*sin^2u*cosu}{2*\sqrt{1-sin^2u} }du=\int\frac{4*sin^2u*cosu}{cosu}du=\\\\ =4*\int sin^2u \ du=4*\int \frac{1-cos2u}{2}du=2*\int(1-cos2u)du=2*\int du- 2*\int cos2u \ du=2*u-sin2u=2*u-2*sinu*cosu.[/tex]
Обратная тригонометрическая подстановка:
[tex]\displaystyle\\2*u-2*sinu*cosu=2*arcsin\frac{x}{2} -2*\frac{x}{2} *\sqrt{1-\frac{x^2}{4} }=2*arcsin \frac{x}{2}-x*\sqrt{1-\frac{x^2}{4} } .\\\\\int\limits^1_0 {\frac{x^2}{\sqrt{4-x^2} } } \, dx =2*arcsin\frac{x}{2} \ |^1_0-x*\sqrt{1-\frac{x^2}{4} } \ |^1_0=\\\\=2*arcsin\frac{1}{2} -2*arcsin0-(1*\sqrt{1-\frac{1^2}{4} }-0*\sqrt{1-\frac{0^2}{4} } }) =\\\\=2*\frac{\pi }{6} -0-(\sqrt{\frac{3}{4} } -0)=\frac{\pi }{3} -\frac{\sqrt{3} }{2} .[/tex]