Answer: The Integral That Sums to an Arcsine

Key Idea / Intuition

The weight $1/\sqrt{x(1-x)}$ is the arclength element of the Beta distribution, and the substitution $x = \sin^2\theta$ transforms the integral into one involving $\ln(\sin^2\theta)$ against a flat measure on $[0,\pi/2]$ — which is precisely twice the classical log-sine integral $\int_0^{\pi/2}\ln(\sin\theta)\,d\theta = -\frac{\pi}{2}\ln 2$. The answer turns out to be a clean multiple of $\pi\ln 2$.

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Formal Proof / Solution

Step 1: Substitution $x = \sin^2\theta$.

Let $x = \sin^2\theta$, so $dx = 2\sin\theta\cos\theta\,d\theta$, and: $\sqrt{x(1-x)} = \sqrt{\sin^2\theta\cos^2\theta} = \sin\theta\cos\theta.$

When $x=0$, $\theta=0$; when $x=1$, $\theta=\pi/2$. Thus:

$I = \int_0^{\pi/2} \frac{\ln(\sin^2\theta)}{\sin\theta\cos\theta}\cdot 2\sin\theta\cos\theta\,d\theta = 2\int_0^{\pi/2}\ln(\sin^2\theta)\,d\theta.$

Step 2: Simplify.

$I = 2\int_0^{\pi/2} 2\ln(\sin\theta)\,d\theta = 4\int_0^{\pi/2}\ln(\sin\theta)\,d\theta.$

Step 3: Use the classical log-sine integral.

The classical result (provable by the reflection symmetry $\theta \mapsto \pi/2 - \theta$ and Wallis's product) states: $\int_0^{\pi/2}\ln(\sin\theta)\,d\theta = -\frac{\pi}{2}\ln 2.$

Step 4: Conclude.

$I = 4\cdot\left(-\frac{\pi}{2}\ln 2\right) = -2\pi\ln 2.$

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Quick verification of the classical log-sine integral (for completeness):

Let $L = \int_0^{\pi/2}\ln(\sin\theta)\,d\theta$. By $\theta\mapsto\pi/2-\theta$, also $L = \int_0^{\pi/2}\ln(\cos\theta)\,d\theta$. Thus: $2L = \int_0^{\pi/2}\ln(\sin\theta\cos\theta)\,d\theta = \int_0^{\pi/2}\ln\!\left(\tfrac{\sin 2\theta}{2}\right)d\theta.$

Substituting $\phi = 2\theta$ and using the $\pi/2$-periodicity: $2L = \tfrac{1}{2}\int_0^{\pi}\ln(\sin\phi)\,d\phi - \tfrac{\pi}{2}\ln 2 = L - \tfrac{\pi}{2}\ln 2,$

so $L = -\tfrac{\pi}{2}\ln 2$. $\checkmark$

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Final answer:

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