Answer: Integral of ln(sin x) via Symmetry

Key Idea / Intuition

The trick is to use a symmetry duplication: pair $\ln(\sin x)$ with $\ln(\cos x)$ by the substitution $x \mapsto \pi/2 - x$, which shows they contribute equally. Then invoke the double-angle identity $\sin(2x) = 2\sin x \cos x$ to reduce the integral to a rescaled copy of itself plus $\ln 2$. Solving the resulting equation gives $I$ exactly.

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Formal Proof / Solution

Step 1: Symmetry.

Let $J = \int_0^{\pi/2} \ln(\cos x)\,dx$. Substituting $x \mapsto \frac{\pi}{2} - x$ gives $J = \int_0^{\pi/2} \ln(\sin x)\,dx = I$. So $I = J$.

Step 2: Add $I + J$.

$2I = I + J = \int_0^{\pi/2} \ln(\sin x)\,dx + \int_0^{\pi/2}\ln(\cos x)\,dx = \int_0^{\pi/2} \ln(\sin x \cos x)\,dx.$

Step 3: Apply the double-angle identity.

$\sin x \cos x = \frac{\sin 2x}{2},$

so

$2I = \int_0^{\pi/2} \ln\!\left(\frac{\sin 2x}{2}\right)dx = \int_0^{\pi/2} \ln(\sin 2x)\,dx - \int_0^{\pi/2}\ln 2\,dx.$

The second integral is $\frac{\pi}{2}\ln 2$.

Step 4: Substitute $u = 2x$ in the first part.

$\int_0^{\pi/2} \ln(\sin 2x)\,dx = \frac{1}{2}\int_0^{\pi}\ln(\sin u)\,du.$

But $\ln(\sin u)$ is symmetric about $u = \pi/2$ on $[0,\pi]$, so

$\frac{1}{2}\int_0^{\pi}\ln(\sin u)\,du = \frac{1}{2}\cdot 2\int_0^{\pi/2}\ln(\sin u)\,du = I.$

Step 5: Solve.

$2I = I - \frac{\pi}{2}\ln 2 \implies I = -\frac{\pi}{2}\ln 2.$

$\boxed{I = \int_0^{\pi/2}\ln(\sin x)\,dx = -\frac{\pi}{2}\ln 2.}$

Why beautiful? The integral appears to depend on a transcendental function in a complicated way, yet the answer is a clean multiple of $\ln 2$. The entire computation reduces to an algebraic equation in $I$ — no special functions, no residues, just a slick symmetry argument.