Use the indicated substitution to evaluate the integral.

${\displaystyle \underset{0}{\overset{\frac{15}{2}}{}}}\frac{x^2}{\sqrt[\phantom{2}]{225-x^2}}dx,x=15sin()$

$($Give an exact answer. Use symbolic notation and fractions where needed.$)$

${\displaystyle \underset{0}{\overset{152}{}}}\frac{x^2}{\sqrt[\phantom{2}]{225-x^2}}dx=$