Answer
$\frac{1}{8}\left( {x\left( {2{x^2} - 1} \right)\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x} \right) + C$
Work Step by Step
$$\eqalign{
& \int {{x^2}\sqrt {1 - {x^2}} } dx \cr
& {\text{Integrate by using a CAS }}\left( {{\text{See image below}}} \right){\text{ we obtain:}} \cr
& \int {{x^2}\sqrt {1 - {x^2}} } dx = \frac{1}{8}\left( {{{\sin }^{ - 1}}x + x\left( {2{x^2} - 1} \right)\sqrt {1 - {x^2}} } \right) + C \cr
& \cr
& {\text{Integrating by tables using the entry 31}} \cr
& \int {{u^2}\sqrt {{a^2} - {u^2}} } du = \frac{u}{8}\left( {2{u^2} - {a^2}} \right)\sqrt {{a^2} - {u^2}} + \frac{{{a^4}}}{8}{\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& {\text{Therefore}}{\text{, let }}a = 1,{\text{ }}u = x \cr
& \int {{x^2}\sqrt {1 - {x^2}} } dx = \frac{x}{8}\left( {2{x^2} - 1} \right)\sqrt {1 - {x^2}} + \frac{1}{8}{\sin ^{ - 1}}x + C \cr
& {\text{Factoring out }}\frac{1}{8} \cr
& \int {{x^2}\sqrt {1 - {x^2}} } dx = \frac{1}{8}\left( {x\left( {2{x^2} - 1} \right)\sqrt {1 - {x^2}} + {{\sin }^{ - 1}}x} \right) + C \cr} $$
