Answer
$$\frac{{3\pi }}{{8\sqrt 2 }}$$
Work Step by Step
$$\eqalign{
& \int_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {{{\left( {1 - 2{x^2}} \right)}^{3/2}}} dx \cr
& \cr
& {\text{write the integrand in terms of }}u \cr
& u = \sqrt 2 x,\,\,\,\,du = \sqrt 2 dx \cr
& \,\,\,\,x = 1/\sqrt 2 \to u = 1{\text{ and }}x = - 1/\sqrt 2 \to u = - 1 \cr
& {\text{then}} \cr
& \int_{ - 1/\sqrt 2 }^{1/\sqrt 2 } {{{\left( {1 - 2{x^2}} \right)}^{3/2}}} dx = \frac{1}{{\sqrt 2 }}\int_{ - 1}^1 {{{\left( {1 - u} \right)}^{3/2}}} du \cr
& \cr
& \cr
& {\text{Use the Endpaper Integral Table to evaluate the integral }} \cr
& {\text{Integrals containing }}{\left( {{a^2} + {u^2}} \right)^{3/2}}{\text{. By the formula 99}} \cr
& \left( {99} \right):{\int {\left( {{a^2} - {u^2}} \right)} ^{3/2}}du = - \frac{u}{8}\left( {2{u^2} - 5{a^2}} \right)\sqrt {{a^2} - {u^2}} + \frac{{3{a^4}}}{8}{\sin ^{ - 1}}\left( {\frac{u}{a}} \right) + C \cr
& \frac{1}{{\sqrt 2 }}{\int {\left( {1 - {u^2}} \right)} ^{3/2}}du = - \frac{u}{{8\sqrt 2 }}\left( {2{u^2} - 5} \right)\sqrt {1 - {u^2}} + \frac{3}{{8\sqrt 2 }}{\sin ^{ - 1}}\left( u \right) + C \cr
& \frac{1}{{\sqrt 2 }}\int_{ - 1}^1 {{{\left( {1 - u} \right)}^{3/2}}} du = \left[ { - \frac{u}{{8\sqrt 2 }}\left( {2{u^2} - 5} \right)\sqrt {1 - {u^2}} + \frac{3}{{8\sqrt 2 }}{{\sin }^{ - 1}}\left( u \right)} \right]_{ - 1}^1 \cr
& \cr
& {\text{evaluate the limits and simplify}} \cr
& = \left[ { - \frac{1}{{8\sqrt 2 }}\left( {2 - 5} \right)\sqrt {1 - {{\left( 1 \right)}^2}} + \frac{3}{{8\sqrt 2 }}{{\sin }^{ - 1}}\left( 1 \right)} \right] - \left[ { - \frac{1}{{8\sqrt 2 }}\left( {2 - 5} \right)\sqrt {1 - {{\left( { - 1} \right)}^2}} + \frac{3}{{8\sqrt 2 }}{{\sin }^{ - 1}}\left( { - 1} \right)} \right] \cr
& = \frac{3}{{8\sqrt 2 }}{\sin ^{ - 1}}\left( 1 \right) - \frac{3}{{8\sqrt 2 }}{\sin ^{ - 1}}\left( { - 1} \right) \cr
& = \frac{{3\pi }}{{16\sqrt 2 }} + \frac{{3\pi }}{{16\sqrt 2 }} \cr
& = \frac{{3\pi }}{{8\sqrt 2 }} \cr} $$
