Answer
$$\frac{{8\sqrt 2 }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^\pi {{{\left( {1 - \cos 2x} \right)}^{3/2}}dx} \cr
& {\text{Use the identity }}\cos 2x = 1 - 2{\sin ^2}x \cr
& \int_0^\pi {{{\left( {1 - \cos 2x} \right)}^{3/2}}dx} = \int_0^\pi {{{\left( {1 - 1 + 2{{\sin }^2}x} \right)}^{3/2}}dx} \cr
& = \int_0^\pi {{{\left( {2{{\sin }^2}x} \right)}^{3/2}}dx} \cr
& = \int_0^\pi {{2^{3/2}}{{\sin }^3}xdx} \cr
& = {2^{3/2}}\int_0^\pi {{{\sin }^2}x\sin xdx} \cr
& {\text{use the pythagorean identity }}{\sin ^2}x + {\cos ^2}x = 1 \cr
& = {2^{3/2}}\int_0^\pi {\left( {1 - {{\cos }^2}x} \right)\sin xdx} \cr
& = {2^{3/2}}\int_0^\pi {\left( {\sin x - {{\cos }^2}x\sin x} \right)dx} \cr
& {\text{Integrate}} \cr
& {\text{ = 2}}\sqrt 2 \left[ { - \cos x + \frac{{{{\cos }^3}x}}{3}} \right]_0^\pi \cr
& = 2\sqrt 2 \left[ {\frac{{{{\cos }^3}\pi }}{3} - \cos \pi } \right] - 2\sqrt 2 \left[ {\frac{{{{\cos }^3}0}}{3} - \cos 0} \right] \cr
& = 2\sqrt 2 \left[ { - \frac{1}{3} + 1} \right] - 2\sqrt 2 \left[ {\frac{1}{3} - 1} \right] \cr
& = - \frac{{2\sqrt 2 }}{3} + 2\sqrt 2 - \frac{{2\sqrt 2 }}{3} + 2\sqrt 2 \cr
& = \frac{{8\sqrt 2 }}{3} \cr} $$
