Espero te ayude:))))))
Respuesta:
[tex]-\frac{10}{9}(1-t^3)+C[/tex]
Explicación paso a paso:
[tex]\displaystyle\int 5t^2\sqrt{1-t^3}~dt =5\displaystyle\int \sqrt{1-t^3}~*t^2dt\\\\1-t^3=u\implies -3t^2dt=du \implies~t^2dt=-\frac{1}{3} {}du\\\\5\displaystyle\int \sqrt{u}*(-\frac{1}{3})du=-\frac{5}{3}\displaystyle\int u ^{\frac{1}{2} } du =-\frac{5}{3}u^{\frac{1}{2}^+^1 } :\frac{3}{2} +C=-\frac{10}{9} u^{\frac{3}{2} }+C}=-\frac{10}{9} u^{\frac{3}{2} } +C[/tex]
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Verified answer
Espero te ayude:))))))
Respuesta:
[tex]-\frac{10}{9}(1-t^3)+C[/tex]
Explicación paso a paso:
[tex]\displaystyle\int 5t^2\sqrt{1-t^3}~dt =5\displaystyle\int \sqrt{1-t^3}~*t^2dt\\\\1-t^3=u\implies -3t^2dt=du \implies~t^2dt=-\frac{1}{3} {}du\\\\5\displaystyle\int \sqrt{u}*(-\frac{1}{3})du=-\frac{5}{3}\displaystyle\int u ^{\frac{1}{2} } du =-\frac{5}{3}u^{\frac{1}{2}^+^1 } :\frac{3}{2} +C=-\frac{10}{9} u^{\frac{3}{2} }+C}=-\frac{10}{9} u^{\frac{3}{2} } +C[/tex]
[tex]-\frac{10}{9}(1-t^3)+C[/tex]