Answer: Integral of ln(sin x)

Key Idea / Intuition

The trick is a beautiful symmetry + duplication argument. Write $I$ using the substitution $x \mapsto \pi/2 - x$ to get $I = \int_0^{\pi/2} \ln(\cos x)\,dx$ as well. Then add the two copies of $I$ together and use the product-to-sum identity $\sin x \cos x = \frac{1}{2}\sin(2x)$. A single substitution then reduces the resulting integral back to $I$ itself, giving a clean equation to solve.

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Formal Proof / Solution

Step 1: Symmetry.

By the substitution $x \mapsto \tfrac{\pi}{2} - x$,

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

Step 2: Add the two copies.

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

Use the identity $\sin x \cos x = \dfrac{\sin 2x}{2}$:

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

Step 3: Substitute $u = 2x$ in the first piece.

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

Now use the symmetry of $\sin$ about $\pi/2$:

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

So $\displaystyle\int_0^{\pi/2} \ln(\sin 2x)\,dx = \frac{1}{2}\cdot 2I = I$.

Step 4: Solve for $I$.

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

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

Result:

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

Why it's surprising: The integral of a function that diverges to $-\infty$ at the endpoint $x=0$ gives a clean closed form involving $\pi$ and $\ln 2$ — two of the most fundamental constants in mathematics, linked here by a symmetry argument rather than any residue or special-function machinery.

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Written to: questions/2025-07-18-PM.md | Answer: questions/2025-07-18-PM-answer.md