Evaluate ∫(0 to π/2) log(sin x) dx — JEE Advanced level definite integral

Question

Evaluate the definite integral:

\int_0^{\pi/2} \log(\sin x) \, dx

(JEE Advanced 2019, similar pattern)

Solution — Step by Step

We know the classic property: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ .

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

Replace $x$ with $\pi/2 - x$ :

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

So $I$ equals both $\int_0^{\pi/2} \log(\sin x)\,dx$ and $\int_0^{\pi/2} \log(\cos x)\,dx$ . This symmetry is the key.

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

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

Using the identity $\sin x \cos x = \frac{\sin 2x}{2}$ :

2I = \int_0^{\pi/2} \log\left(\frac{\sin 2x}{2}\right)dx

2I = \int_0^{\pi/2} \log(\sin 2x)\,dx - \int_0^{\pi/2} \log 2\,dx

The second integral is straightforward:

\int_0^{\pi/2} \log 2\,dx = \frac{\pi}{2}\log 2

For the first integral, substitute $t = 2x$ , so $dt = 2\,dx$ :

\int_0^{\pi/2} \log(\sin 2x)\,dx = \frac{1}{2}\int_0^{\pi} \log(\sin t)\,dt

Since $\sin(\pi - t) = \sin t$ , the function $\log(\sin t)$ is symmetric about $t = \pi/2$ on $[0, \pi]$ .

\frac{1}{2}\int_0^{\pi} \log(\sin t)\,dt = \frac{1}{2} \cdot 2\int_0^{\pi/2} \log(\sin t)\,dt = I

So we get: $2I = I - \frac{\pi}{2}\log 2$

2I - I = -\frac{\pi}{2}\log 2

\boxed{I = -\frac{\pi}{2}\log 2}

The negative sign makes sense — $\sin x \leq 1$ on $[0, \pi/2]$ , so $\log(\sin x) \leq 0$ throughout the interval.

Why This Works

The entire solution hinges on a symmetry trick: replacing $x$ with $\pi/2 - x$ converts $\sin$ to $\cos$ , giving us two forms of the same integral. Adding them brings in the double-angle identity, which introduces a substitution that loops back to the original integral $I$ .

This “self-referencing” technique — where manipulating $I$ produces an equation in $I$ itself — is a hallmark of definite integral problems at the JEE Advanced level. The integral cannot be computed by finding an antiderivative (there is no closed-form antiderivative of $\log(\sin x)$ ), so these symmetry-based tricks are the only way forward.

Alternative Method — Using Fourier Series

For those comfortable with series, $\log(\sin x)$ has the Fourier expansion:

\log(\sin x) = -\log 2 - \sum_{k=1}^{\infty} \frac{\cos 2kx}{k}

Integrating term by term from $0$ to $\pi/2$ :

I = -\frac{\pi}{2}\log 2 - \sum_{k=1}^{\infty} \frac{1}{k}\int_0^{\pi/2}\cos 2kx\,dx

Each integral $\int_0^{\pi/2}\cos 2kx\,dx = \frac{\sin(k\pi)}{2k} = 0$ for all positive integers $k$ .

So $I = -\frac{\pi}{2}\log 2$ . Same answer, confirmed.

This integral is a standard result that appears frequently in JEE Advanced. Memorise the answer: $\int_0^{\pi/2} \log(\sin x)\,dx = -\frac{\pi}{2}\ln 2$ . It also shows up as a sub-step in harder integrals involving $\log(\sin x + \cos x)$ or $\log(\tan x)$ .

Common Mistake

Students often try to find an antiderivative of $\log(\sin x)$ using integration by parts. This leads to $\int \cot x\,dx$ nested inside further integrals, creating an endless loop. The function $\log(\sin x)$ does not have a closed-form antiderivative — you must use the definite integral properties (symmetry, King’s rule) to evaluate it. If you find yourself doing IBP on this problem, stop and switch strategy.

Question

Solution — Step by Step

Why This Works

Alternative Method — Using Fourier Series

Common Mistake

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