Step 1: Use $ \cos^2\theta=\frac{1+\cos 2\theta}{2} $.
Let $ A=\frac{3\pi}{8}-\frac{x}{4} $, $ B=\frac{11\pi}{8}+\frac{x}{4} $. Then: \[ \cos^2 A-\cos^2 B=\frac{\cos 2A-\cos 2B}{2} \]
Step 2: Find $ 2A=\frac{3\pi}{4}-\frac{x}{2} $ and $ 2B=\frac{11\pi}{4}+\frac{x}{2} $.
Step 3: Apply $ \cos P-\cos Q=-2\sin\frac{P+Q}{2}\sin\frac{P-Q}{2} $.
$ \frac{P+Q}{2}=\frac{7\pi}{4} $ and $ \frac{P-Q}{2}=-\pi-\frac{x}{2} $.
Step 4: Simplify.
$ \sin(7\pi/4)=-\frac{1}{\sqrt{2}} $ and $ \sin(-\pi-x/2)=\sin(x/2) $. So: \[ \cos 2A-\cos 2B=-2\cdot\left(-\frac{1}{\sqrt{2}}\right)\sin\frac{x}{2}=\sqrt{2}\sin\frac{x}{2} \]
Step 5: Evaluate the integral.
\[ I=\int_0^\pi\frac{\sqrt{2}\sin(x/2)}{2}dx=\frac{\sqrt{2}}{2}\left[-2\cos\frac{x}{2}\right]_0^\pi=\frac{\sqrt{2}}{2}\cdot(-2)(0-1)=\sqrt{2} \]
Step 6: State the answer.
\[ \boxed{\sqrt{2}} \]