Answer
$\frac{\pi }{8}$
Work Step by Step
$$\eqalign{
&\text{Let }I= \int_0^{\pi /4} {{{\sin }^2}\left( {2\theta } \right)} d\theta \cr
& {\text{Use the identity si}}{{\text{n}}^2}x = \frac{{1 - \cos 2x}}{2},{\text{ let }}x = 2\theta \cr
&I= \int_0^{\pi /4} {\frac{{1 - \cos 2\left( {2\theta } \right)}}{2}} d\theta \cr
& = \int_0^{\pi /4} {\frac{{1 - \cos 4\theta }}{2}} d\theta \cr
& = \int_0^{\pi /4} {\left( {\frac{1}{2} - \frac{{\cos 4\theta }}{2}} \right)} d\theta \cr
& = \frac{1}{2}\int_0^{\pi /4} {\left( {1 - \cos 4\theta } \right)} d\theta \cr
& {\text{Integrate}}{\text{, use }}\int {\cos ax} dx = \frac{1}{a}\sin ax + C \cr
& I= \frac{1}{2}\left[ {\theta - \frac{1}{4}\sin 4\theta } \right]_0^{\pi /4} \cr
& {\text{Evaluate}} \cr
& I= \frac{1}{2}\left[ {\frac{\pi }{4} - \frac{1}{4}\sin 4\left( {\frac{\pi }{4}} \right)} \right] - \frac{1}{2}\left[ {0 - \frac{1}{4}\sin 4\left( 0 \right)} \right] \cr
& = \frac{1}{2}\left[ {\frac{\pi }{4} - \frac{1}{4}\left( 0 \right)} \right] - \frac{1}{2}\left[ 0 \right] \cr
& = \frac{\pi }{8} \cr} $$
