[tex]\int\limits^2_0 {x^3\sqrt{4-x^2} } \, dx\\ u=\sqrt{4-x^2}|()^2=>u^2=4-x^2=>x^2=4-u^2=>x=\sqrt{4-u^2} \\dx=-\frac{u}{\sqrt{4-u^2}} du\\x^3=(\sqrt{4-u^2})^3\\x=0=>u=2\\x=2=>u=0\\\int\limits^0_2 {x^3\sqrt{4-x^2} } \, dx=\int\limits^0_2 {-\frac{(\sqrt{4-u^2})^3u*u}{\sqrt{4-u^2}} } \, du=\int\limits^0_2{-(4-u^2)u^2 } \, du=\int\limits^0_2 {(u^4-4u^2 )} \, du=[/tex]
[tex]=-\int\limits^2_0 {(u^4-4u^2) } \, du=\int\limits^2_0 {u^4} \, du+4\int\limits^2_0{u^2} \, du= -\frac{u^{4+1}}{4+1}|^2_0 +4*\frac{u^2+1}{2+1}|^2_0= - \frac{u^5}{5}|^2_0+4*\frac{u^3}{3}|^2_0=-\frac{2^5}{5}+\frac{0^5}{5}+4*\frac{2^3}{3}-4*\frac{0^3}{3}= -\frac{32}{5}+\frac{32}{3}=\frac{-96+160}{15}=\frac{64}{15}\\\int\limits^2_0{x^3\sqrt{4-x^3} } \, dx =\frac{64}{15}[/tex]
