Answer
\[ = \frac{1}{{\sqrt 2 }}\,\left( {1 - \frac{1}{{\sqrt 2 }}} \right)\]
Work Step by Step
\[\begin{gathered}
\int\limits_0^{\frac{\pi }{8}} {\sqrt {1 - \cos 4x} \,dx} \hfill \\
\hfill \\
use\,\,the\;\;identities \hfill \\
\hfill \\
1 = {\sin ^2}2x + {\cos ^2}2x\,\,\,\,and\,\,\,\,\cos \,4x = {\cos ^2}2x - {\sin ^2}2x \hfill \\
\hfill \\
therefore \hfill \\
\hfill \\
= \int\limits_0^{\frac{\pi }{8}} {\sqrt {{{\sin }^2}2x + {{\cos }^2}2x - {{\cos }^2}2x + {{\sin }^2}2x} \,\,dx} \hfill \\
\hfill \\
simplify\,\,inside\,\,the\,\,radical \hfill \\
\hfill \\
= \sqrt 2 \int\limits_0^{\frac{\pi }{8}} {\sin 2x\,dx} \hfill \\
\hfill \\
set\,\,\,\,\,2x = t\,\,\, \to \,\,2dx = dt\,\,and\,\,dx = \frac{{dt}}{2} \hfill \\
\hfill \\
= \frac{{\sqrt 2 }}{2}\,\int\limits_0^{\frac{\pi }{4}} {\sin t\,dt} \hfill \\
\hfill \\
\,integrate\,and\,use\,\,the\,\,ftc \hfill \\
\hfill \\
= \frac{1}{{\sqrt 2 }}\,\,\left[ { - \cos t} \right]_0^{\frac{\pi }{4}} \hfill \\
\hfill \\
= \frac{{ - 1}}{{\sqrt 2 }}\,\left( {\frac{1}{{\sqrt 2 }} - 1} \right) \hfill \\
\hfill \\
= \frac{1}{{\sqrt 2 }}\,\left( {1 - \frac{1}{{\sqrt 2 }}} \right) \hfill \\
\hfill \\
\end{gathered} \]
