Evaluate using property ∫₀^a f(x)dx = ∫₀^a f(a-x)dx. Result: -(π/2)log 2.

∫₀^π/2 log(sin x) dx — Famous Definite Integral

Question

Evaluate the definite integral:

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

This is a JEE Advanced favourite — it appears almost every 3-4 years in some form. The trick is not computation; it’s recognizing which property to apply.

Solution — Step by Step

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

We use the reflection property: $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ . With $a = \pi/2$ , replacing $x$ with $(\pi/2 - x)$ :

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

We now have two equal-looking integrals. Adding them:

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

Write $\sin x \cos x = \frac{\sin 2x}{2}$ . The log splits:

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

The second integral is trivial: $\int_0^{\pi/2} \log 2\, dx = \frac{\pi}{2}\log 2$ .

For $J = \int_0^{\pi/2} \log(\sin 2x)\, dx$ , substitute $t = 2x$ , so $dt = 2\,dx$ :

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

Now use the property $\int_0^{\pi} \log(\sin t)\, dt = 2\int_0^{\pi/2} \log(\sin t)\, dt$ . (This follows because $\sin(\pi - t) = \sin t$ , so the function is symmetric about $\pi/2$ .)

J = \frac{1}{2} \cdot 2I = I

Substituting back:

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

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

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

Why This Works

The reflection property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ doesn’t change the value of the integral — it just relabels where we’re integrating from. For $a = \pi/2$ , it converts $\sin x$ to $\cos x$ , which is the key move here.

Adding the two forms lets us combine $\log \sin + \log \cos$ into $\log(\sin 2x/2)$ , which creates a self-referential equation: the new integral $J$ turns out to equal $I$ itself after a substitution. This “I appears on both sides” technique is the hallmark of this entire class of problems.

The result is negative because $\log(\sin x) < 0$ for all $x \in (0, \pi/2)$ — $\sin x$ lives strictly between 0 and 1 there, and $\log$ of a number less than 1 is negative. So a negative answer is physically expected.

Alternative Method

For NEET or board-level: this integral can also be remembered as a standard result — $\int_0^{\pi/2} \log(\sin x)\, dx = \int_0^{\pi/2} \log(\cos x)\, dx = -\frac{\pi}{2}\log 2$ . Both integrals give the same value (which Step 1 proves directly).

Some JEE problems present this in disguise by asking $\int_0^{\pi/2} \log(\cos x)\, dx$ or even $\int_0^{\pi} \log(\sin x)\, dx$ . For the latter, use the symmetry argument from Step 4: $\int_0^{\pi} \log(\sin x)\, dx = 2I = -\pi \log 2$ .

Common Mistake

The most frequent error is misapplying the symmetry in Step 4. Students write $\int_0^{\pi} \log(\sin t)\, dt = \int_0^{\pi/2} \log(\sin t)\, dt$ (forgetting the factor of 2). The function $\sin t$ is symmetric about $t = \pi/2$ on $[0, \pi]$ , which means the integral over $[0, \pi]$ is twice the integral over $[0, \pi/2]$ . Missing this factor of 2 gives $I = -\frac{\pi}{4}\log 2$ — half the correct answer.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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