$\int_{0}^{\frac{\pi}{8}} \sin 2 x d x$

Step 1 ：The problem is to evaluate the integral of \(\sin 2x\) from 0 to \(\frac{\pi}{8}\).

Step 2 ：The integral of \(\sin 2x\) is \(-\frac{1}{2}\cos 2x\).

Step 3 ：We need to evaluate this from 0 to \(\frac{\pi}{8}\) and subtract the two results.

Step 4 ：The result of the integral is \(\frac{1}{2} - \frac{\sqrt{2}}{4}\).

Step 5 ：Final Answer: \(\boxed{\frac{1}{2} - \frac{\sqrt{2}}{4}}\)