Answer
$$\frac{{2\sqrt 2 }}{3}$$
Work Step by Step
$$\eqalign{
& \int_0^{\pi /4} {{{\left( {1 + \cos 4x} \right)}^{3/2}}dx} \cr
& = \int_0^{\pi /4} {{{\left[ {1 + \cos 2\left( {2x} \right)} \right]}^{3/2}}dx} \cr
& {\text{Use the identity }}\cos 2\theta = 2{\cos ^2}\left( {2\theta } \right) - 1 \cr
& \int_0^{\pi /4} {{{\left[ {1 + \cos 2\left( {2x} \right)} \right]}^{3/2}}dx} = \int_0^{\pi /4} {{{\left[ {1 + 2{{\cos }^2}\left( {2x} \right) - 1} \right]}^{3/2}}dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{\pi /4} {{{\left[ {2{{\cos }^2}\left( {2x} \right)} \right]}^{3/2}}dx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \int_0^{\pi /4} {{2^{3/2}}{{\cos }^3}2xdx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2^{3/2}}\int_0^{\pi /4} {\left( {1 - {{\sin }^2}2x} \right)\cos 2xdx} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = {2^{3/2}}\int_0^{\pi /4} {\left( {\cos 2x - {{\sin }^2}2x\cos 2x} \right)dx} \cr
& {\text{Integrate}} \cr
& = {2^{3/2}}\left[ {\frac{1}{2}\sin 2x - \frac{1}{6}{{\sin }^3}2x} \right]_0^{\pi /4} \cr
& = {2^{3/2}}\left[ {\frac{1}{2}\sin \left( {\frac{\pi }{2}} \right) - \frac{1}{6}{{\sin }^3}\left( {\frac{\pi }{2}} \right)} \right] - {2^{3/2}}\left[ {\frac{1}{2}\sin \left( 0 \right) - \frac{1}{6}{{\sin }^3}\left( 0 \right)} \right] \cr
& = {2^{3/2}}\left[ {\frac{1}{2} - \frac{1}{6}} \right] \cr
& = \frac{{2\sqrt 2 }}{3} \cr} $$
