Answer
$\displaystyle \frac{\sqrt{2}}{2}$
Work Step by Step
Use the identity $\cos 2x=1-2\sin^{2}x$
$ I=\displaystyle \int_{0}^{\pi/4}\sqrt{1-\cos 4\theta}d\theta=\int_{0}^{\pi/4}\sqrt{1-(1-2\sin^{2}2\theta)}d\theta$
$=\displaystyle \int_{0}^{\pi/4}\sqrt{2\sin^{2}2\theta}d\theta$
$=\displaystyle \sqrt{2}\int_{0}^{\pi/4}|\sin 2\theta|d\theta$
... sine is positive on the interval $[0,\pi/2]$
... so we can lose the absolute value brackets
$=\displaystyle \sqrt{2}[-\frac{1}{2}\cos 2\theta]_{0}^{\pi/4}$
$=-\displaystyle \frac{\sqrt{2}}{2}(0-1)$
$=\displaystyle \frac{\sqrt{2}}{2}$
