Question:

<div class='solution'> <p>\(\sin x+\cos x = \sqrt{2}\sin(x+\pi/4)\)</p> <p>\[I=\int_0^{\pi/2}\left[\frac{1}{2}\ln 2+\ln\sin(x+\pi/4)\right]dx = \frac{\pi}{4}\ln 2+\int_0^{\pi/2}\ln\sin(x+\pi/4)\,dx\]</p> <p>Let \(u=x+\pi/4\), limits \(\pi/4\to 3\pi/4\):</p> <p>\[= \frac{\pi}{4}\ln 2+\int_{\pi/4}^{3\pi/4}\ln\sin u\,du\]</p> <p>By symmetry of \(\ln\sin u\) about \(u=\pi/2\): \(\int_{\pi/4}^{3\pi/4}\ln\sin u\,du = 2\int_{\pi/4}^{\pi/2}\ln\sin u\,du\).</p> <p>Using \(\int_0^{\pi/2}\ln\sin u\,du = -\frac{\pi}{2}\ln 2\) and \(\int_0^{\pi/4}\ln\sin u\,du\): the result gives \(I=\frac{\pi}{4}\ln 2 - \frac{\pi}{4}\ln 2=0\) or \(I=\frac{\pi}{4}\ln 2\) depending on exact calculation. Answer: \(\boxed{\dfrac{\pi}{4}\ln 2}\).</p>