Prove the Integral of log sin x from 0 to pi/2

This lesson proves the standard result that $\int_0^{\pi/2} \log(\sin x)\,dx = -\tfrac{\pi}{2}\log 2$. It is both a worked proof and a result worth remembering, since the same value appears in many exam questions. The whole proof rests on the properties of definite integrals.

Setting up the integral

Let

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

This is equation (1).

Applying the king property

Use the property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ with $a = \tfrac{\pi}{2}$. Replacing $x$ by $\tfrac{\pi}{2} - x$ and using $\sin\!\left(\tfrac{\pi}{2} - x\right) = \cos x$ gives

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

This is equation (2).

Adding the two forms

Adding equations (1) and (2):

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

using $\log a + \log b = \log(ab)$.

The log 2 trick

To turn $\sin x \cos x$ into $\sin 2x$, add and subtract $\log 2$:

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

Since $2\sin x \cos x = \sin 2x$, and the second integral is a constant times the length of the interval,

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

Call this equation (3).

Evaluating the remaining integral

Let

$I_1 = \int_0^{\pi/2} \log(\sin 2x)\,dx.$

Substitute $2x = t$, so $2\,dx = dt$ and $dx = \tfrac{1}{2}\,dt$. When $x = 0$, $t = 0$; when $x = \tfrac{\pi}{2}$, $t = \pi$. Therefore

$I_1 = \frac{1}{2}\int_0^{\pi} \log(\sin t)\,dt.$

Folding back with symmetry

Apply the property $\int_0^{2a} f(x)\,dx = 2\int_0^{a} f(x)\,dx$ when $f(2a - x) = f(x)$. Here $2a = \pi$, and $\sin(\pi - t) = \sin t$, so $\log(\sin(\pi - t)) = \log(\sin t)$ and the condition holds. Thus

$I_1 = \frac{1}{2}\cdot 2\int_0^{\pi/2} \log(\sin t)\,dt = \int_0^{\pi/2} \log(\sin t)\,dt = I,$

since the variable of integration is just a dummy name.

Solving for I

Put $I_1 = I$ back into equation (3):

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

So $2I - I = -\tfrac{\pi}{2}\log 2$, which gives

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

Key takeaways

• $\int_0^{\pi/2} \log(\sin x)\,dx = \int_0^{\pi/2} \log(\cos x)\,dx = -\tfrac{\pi}{2}\log 2$.
• The king property $\int_0^a f(x)\,dx = \int_0^a f(a-x)\,dx$ turns sine into cosine and lets you add the two forms.
• Adding and subtracting $\log 2$ creates $\sin 2x$, and the symmetry property folds the integral back onto itself so you can solve for $I$ directly.