Step 1: Working with cos instead as the primary variable:
Define \(J=\displaystyle\int_0^{\pi/2}\log\cos x\,dx\); by the same reflection substitution \(x\to\pi/2-x\), \(J=\displaystyle\int_0^{\pi/2}\log\sin x\,dx=I\), so \(I=J\) — the two integrals are automatically equal by symmetry.
Step 2: Adding I and J directly:
\(I+J=2I=\displaystyle\int_0^{\pi/2}\log(\sin x\cos x)dx=\int_0^{\pi/2}\log\left(\dfrac{\sin2x}2\right)dx\).
Step 3: Splitting the logarithm:
\(=\displaystyle\int_0^{\pi/2}\log\sin2x\,dx-\int_0^{\pi/2}\log2\,dx=\int_0^{\pi/2}\log\sin2x\,dx-\dfrac\pi2\log2\).
Step 4: Substituting u=2x and using periodic symmetry of sin:
With \(u=2x\): \(\displaystyle\int_0^{\pi/2}\log\sin2x\,dx=\dfrac12\int_0^\pi\log\sin u\,du\). Splitting \([0,\pi]\) at \(\pi/2\) and using \(\sin(\pi-u)=\sin u\) on the second half shows both halves are equal, so this equals \(\dfrac12\cdot2I=I\).
Step 5: Solving the resulting equation for I:
\(2I=I-\dfrac\pi2\log2\Rightarrow I=-\dfrac\pi2\log2\), confirming the same result via the symmetric cos-based route.
Final Answer:
\[ \boxed{I=-\dfrac{\pi}{2}\log2} \]