Evaluate ∫₀^π/2 ln(sin x) , dx — JEE Mathematics

Let $I = \int_{0}^{\pi/2} \ln(\sin x) \, dx \quad \dots (1)$
Using the property $\int_{0}^{a} f(x) \, dx = \int_{0}^{a} f(a-x) \, dx$:

$$I = \int_{0}^{\pi/2} \ln(\sin(\pi/2 - x)) \, dx = \int_{0}^{\pi/2} \ln(\cos x) \, dx \quad \dots (2)$$

Adding (1) and (2):

$$2I = \int_{0}^{\pi/2} [\ln(\sin x) + \ln(\cos x)] \, dx = \int_{0}^{\pi/2} \ln(\sin x \cos x) \, dx$$

Multiply and divide by 2 inside the logarithm:

$$2I = \int_{0}^{\pi/2} \ln\left(\frac{\sin 2x}{2}\right) \, dx = \int_{0}^{\pi/2} \ln(\sin 2x) \, dx - \int_{0}^{\pi/2} \ln 2 \, dx$$

$$2I = \int_{0}^{\pi/2} \ln(\sin 2x) \, dx - \frac{\pi}{2}\ln 2 \quad \dots (3)$$

For the first sub-integral, let $2x = t \implies dx = \frac{dt}{2}$. Limits change from $0 \to \pi$.

$$\int_{0}^{\pi/2} \ln(\sin 2x) \, dx = \frac{1}{2} \int_{0}^{\pi} \ln(\sin t) \, dt$$

Using the periodic / symmetric property $\int_{0}^{2a} f(t) \, dt = 2\int_{0}^{a} f(t) \, dt$ if $f(2a-t) = f(t)$:

$$\frac{1}{2} \int_{0}^{\pi} \ln(\sin t) \, dt = \frac{1}{2} \cdot 2 \int_{0}^{\pi/2} \ln(\sin t) \, dt = I$$

Substitute this back into equation (3):

$$2I = I - \frac{\pi}{2}\ln 2 \implies I = -\frac{\pi}{2}\ln 2$$

Answer: $-\frac{\pi}{2}\ln 2$