Answer
$$18{\sin ^{ - 1}}\left( {\frac{x}{6}} \right) - \frac{{x\sqrt {36 - {x^2}} }}{2} + C$$
Work Step by Step
$$\eqalign{
& \int {\frac{{{x^2}}}{{\sqrt {36 - {x^2}} }}} dx \cr
& {\text{Refer to the triangle below}} \cr
& \sin \theta = \frac{x}{6} \cr
& x = 6\sin \theta ,{\text{ }}dx = 6\cos \theta d\theta ,{\text{ }} \cr
& {\text{Substituting}} \cr
& \int {\frac{{{x^2}}}{{\sqrt {36 - {x^2}} }}} dx = \int {\frac{{36{{\sin }^2}\theta }}{{\sqrt {36 - {{\left( {6\sin \theta } \right)}^2}} }}} \left( {6\cos \theta } \right)d\theta \cr
& = \int {\frac{{36{{\sin }^2}\theta }}{{\sqrt {36 - 36{{\sin }^2}\theta } }}} \left( {6\cos \theta } \right)d\theta \cr
& = \int {\frac{{36{{\sin }^2}\theta }}{{\sqrt {36\left( {1 - {{\sin }^2}\theta } \right)} }}} \left( {6\cos \theta } \right)d\theta \cr
& = \int {\frac{{36{{\sin }^2}\theta }}{{\sqrt {36{{\cos }^2}\theta } }}} \left( {6\cos \theta } \right)d\theta \cr
& = \int {\frac{{36{{\sin }^2}\theta }}{{6\cos \theta }}} \left( {6\cos \theta } \right)d\theta \cr
& = \int {36{{\sin }^2}\theta d\theta } \cr
& = 36\int {{{\sin }^2}\theta d\theta } \cr
& {\text{Using the power reducing formula si}}{{\text{n}}^2}u = \frac{{1 - \cos 2u}}{2} \cr
& = 36\int {\left( {\frac{1}{2} - \frac{1}{2}\cos 2\theta } \right)} d\theta \cr
& {\text{Integrating}} \cr
& = 36\left( {\frac{1}{2}\theta - \frac{1}{4}\sin 2\theta } \right) + C \cr
& = 36\left( {\frac{1}{2}\theta - \frac{1}{2}\sin \theta \cos \theta } \right) + C \cr
& = 18\theta - 18\sin \theta \cos \theta + C \cr
& {\text{Refer to the triangle }}\cos \theta = \frac{{\sqrt {36 - {x^2}} }}{6},{\text{ }} \cr
& x = 6\sin \theta \to \theta = {\sin ^{ - 1}}\left( {\frac{x}{6}} \right) \cr
& = 18{\sin ^{ - 1}}\left( {\frac{x}{6}} \right) - 18\left( {\frac{x}{6}} \right)\left( {\frac{{\sqrt {36 - {x^2}} }}{6}} \right) + C \cr
& = 18{\sin ^{ - 1}}\left( {\frac{x}{6}} \right) - \frac{{x\sqrt {36 - {x^2}} }}{2} + C \cr} $$
