Q.Evaluate $\int_0^{\pi/2} \log \sin x\, dx$

The key idea is to use the symmetry of the definite integral, often called the King's property: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$.

Let $I = \int_0^{\pi/2} \log \sin x\, dx$.

Using the substitution $x \to \frac{\pi}{2} - x$, we get

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

Add the two expressions for $I$:

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

Use $\sin x \cos x = \frac{1}{2} \sin 2x$, so

$2I = \int_0^{\pi/2} \log\left(\frac{1}{2} \sin 2x\right) dx = \int_0^{\pi/2} \log \frac{1}{2}\, dx + \int_0^{\pi/2} \log \sin 2x\, dx$.

The first term is $-\frac{\pi}{2} \log 2$. For the second, let $t = 2x$, then $dx = dt/2$, limits $0$ to $\pi$: …

The integral $I = \int_0^{\pi/2} \log \sin x \, dx$ is a famous one — it appears in many contexts, from probability to number theory. The trick is not to integrate directly (the antiderivative involves the dilogarithm), but to exploit symmetry.

Why symmetry works: The interval $[0, \pi/2]$ is symmetric about $\pi/4$. The function $\log \sin x$ is not symmetric itself, but if we replace $x$ by $\frac{\pi}{2} - x$, we get $\log \cos x$. Adding the two forms gives $\log(\sin x \cos x) = \log(\frac{1}{2}\sin 2x)$, which splits into a constant term and a scaled version of the original integral. This creates an equation we can solve for $I$.

Let’s walk through it step by step.

1. Define the integral and apply the substitution $x \to \frac{\pi}{2} - x$. Let $I = \int_0^{\pi/2} \log \sin x \, dx$. Substitute $x = \frac{\pi}{2} - t$. Then $dx = -dt$, and when $x=0$, $t=\pi/2$; when $x=\pi/2$, $t=0$. So:

Renaming the dummy variable back to $x$, we have:

So the integral of $\log \sin x$ equals the integral of $\log \cos x$ over the same interval.

2. Add the two expressions for $I$.

Using the identity $\sin x \cos x = \frac{1}{2} \sin 2x$, we get:

The constant $\log(1/2) = -\log 2$ factors out:

3. Handle the integral $\int_0^{\pi/2} \log \sin 2x \, dx$ with another substitution. Let $u = 2x$. Then $dx = du/2$, and when $x=0$, $u=0$; when $x=\pi/2$, $u=\pi$. So:

Now, the integral from $0$ to $\pi$ of $\log \sin u$ can be split at $\pi/2$:

$\int_0^{\pi} \log \sin u \, du = \int_0^{\pi/2} \log \sin u \, du + \int_{\pi/2}^{\pi} \log \sin u \, du.$ …

Method: Reflection plus double-angle self-similarity for $\log$-trig integrals

Use this for integrals like $\int_0^{\pi/2}\log\sin x\,dx$ where direct antidifferentiation fails (the antiderivative is non-elementary) but the interval $[0,\tfrac{\pi}{2}]$ has $\sin\leftrightarrow\cos$ symmetry.

Steps

Step 1: Reflect with $x\to\tfrac{\pi}{2}-x$.

By $\int_0^a f(x)\,dx=\int_0^a f(a-x)\,dx$, the integral of $\log\sin x$ equals the integral of $\log\cos x$ over the same interval. So $I$ can be written two ways.

Step 2: Add the two forms and use a product identity.

Splitting the log separates a constant term $\int_0^{\pi/2}\log\tfrac12\,dx$ from a new $\log\sin 2x$ term.

Step 3: Rescale the double angle back to the original integral. …

Common Mistakes

Mistake 1: Dropping the $\tfrac12$ factor when substituting $t=2x$.

Why it's wrong: $t=2x$ gives $dx=\tfrac{dt}{2}$, so $\int_0^{\pi/2}\log\sin 2x\,dx=\tfrac12\int_0^{\pi}\log\sin t\,dt$; omitting the $\tfrac12$ doubles that term and gives a wrong final value. Correct approach: carry the Jacobian $dx=\tfrac{dt}{2}$ through carefully.

Mistake 2: Splitting $\log(\tfrac12\sin 2x)$ incorrectly. …