Интеграл

[abc]

I= ∫x^3 √(9-4x^2) dx

[Добромир Глухаров]

[tex]\int x^3(9-4x^2)^{\frac{1}{2}}dx[/tex]

[tex]9-4x^2=t^2[/tex]

[tex]x^2=\frac{9-t^2}{4}[/tex]

[tex]x=\frac{1}{2}(9-t^2)^{\frac{1}{2}}[/tex]

[tex]dx=\frac{1}{2}\cdot\frac{1}{2}(9-t^2)^{-\frac{1}{2}}.(-2t)dt=-\frac{1}{2}t(9-t^2)^{-\frac{1}{2}}dt[/tex]

[tex]\int x^3(9-4x^2)^{\frac{1}{2}}dx=\int \frac{1}{8}(9-t^2)^{\frac{3}{2}}.t.\(-\frac{1}{2}\)t(9-t^2)^{-\frac{1}{2}}dt=-\frac{1}{16}\int t^2(9-t^2)dt=\\=-\frac{1}{16}\(\frac{9t^3}{3}-\frac{t^5}{5}\)+C=-\frac{3}{16}t^3+\frac{1}{80}t^5+C=-\frac{3}{16}\sqrt{(9-4x^2)^3}+\frac{1}{80}\sqrt{(9-4x^2)^5}+C[/tex]