Answer
$${\text{True}}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1}^1 {{x^2}\sqrt {1 - {x^2}} dx} \cr
& {\text{Let }}x = \sin \theta ,{\text{ }}dx = \cos \theta d\theta \cr
& {\text{The new limits of integration are}} \cr
& x = - 1,{\text{ }}\theta = {\sin ^{ - 1}}\left( { - 1} \right) = - \frac{1}{2}\pi \cr
& x = 1,{\text{ }}\theta = {\sin ^{ - 1}}\left( 1 \right) = \frac{1}{2}\pi \cr
& {\text{Substituting}} \cr
& \int_{ - 1}^1 {{x^2}\sqrt {1 - {x^2}} dx} = \int_{ - \pi /2}^{\pi /2} {{{\sin }^2}\theta \sqrt {1 - {{\sin }^2}\theta } \cos \theta d\theta } \cr
& = \int_{ - \pi /2}^{\pi /2} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta } \cr
& {\text{By using properties of integrals}} \cr
& = 2\int_0^{\pi /2} {{{\sin }^2}\theta {{\cos }^2}\theta d\theta } \cr
& {\text{The statement is True}} \cr} $$
