Answer
\[\frac{1}{\sqrt 2}\]
Work Step by Step
Considering indefinite integral
Let \[I=\int\sqrt{1-\cos 4\theta}\;d\theta\]
\[I=\int\sqrt{1-\cos 2(2\theta)}\;d\theta\]
\[\left[2\sin^2 x=1-\cos 2x\right]\]
\[I=\int\sqrt{2\sin^2 2\theta}\;d\theta\]
\[I=\sqrt{2}\int\sin 2\theta\;d\theta\]
$I=-\frac{\sqrt{2}}{2}\cos 2\theta=-\frac{1}{\sqrt 2}\cos 2\theta$ ___(1)
Let \[I_1=\int_{0}^{\frac{π}{4}}\sqrt{1-\cos 4\theta}\;d\theta\]
\[I_1=\left[-\frac{1}{\sqrt 2}\cos 2\theta\right]_{0}^{\frac{π}{4}}\]
Using (1)
\[I_1=-\frac{1}{\sqrt 2}\left[\cos \frac{π}{2}-\cos (0)\right]\]
\[I_1=-\frac{1}{\sqrt{2}}[0-1]=\frac{1}{\sqrt 2}\]
Hence, $I_1=\large\frac{1}{\sqrt{2}}$.
