Using symmetry (advanced)

$\begin{aligned} I & = \int_0^\frac \pi 2 \ln (\sin x) \ dx & \small \color{grey} \text{Using }\int_a^b f(x) \ dx = \int_a^b f(a+b-x)\ dx \\ & = \frac 12 \int_0^\frac \pi 2 (\ln (\sin x) + \ln (\cos x))\ dx \\ & = \frac 12 \int_0^\frac \pi 2 \ln (\sin x \cos x)\ dx \\ & = \frac 12 \int_0^\frac \pi 2 \ln \left(\frac 12 \sin (2x) \right) \ dx \\ & = {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\sin (2x))\ dx} - \frac 12 \int_0^\frac \pi 2 \ln 2 \ dx & \small \color{#3D99F6} \text{Let } u = 2x \implies du = 2 \ dx \\ & = {\color{#3D99F6} \frac 14 \int_0^\pi \ln (\sin u)\ du} - \frac {\pi\ln 2}4 & \small \color{#3D99F6} \sin u \text{ is symmetrical about }\frac \pi 2 \\ & = {\color{#3D99F6} \frac 12 \int_0^\frac \pi 2 \ln (\sin u)\ du} - \frac {\pi\ln 2}4 \\ & = {\color{#3D99F6} \frac 12 I} - \frac {\pi\ln 2}4 \\ & = \boxed{-\dfrac {\pi \ln 2}2} \end{aligned}$

Let's start with showing that the hint is correct. Since $\sin(x)$ = $\cos(\pi/2-x)$ for every x between 0 and $\pi/2$ there is a point between 0 and $\pi/2$ for which $\sin(x)$ has an equal cosine. Thus, the integrals are equal.

Using the hint we see that

$\displaystyle\int_{0}^{\pi/2} \ln(\sin(x))dx=\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(x))+\ln(\cos(x))dx=$

$\displaystyle\frac {1}{2}\int_{0}^{\pi/2} \ln(\sin(x)\cos(x))dx$

Now we use the fact that $\sin(x)\cos(x)=\frac{1}{2}(\sin2x)$ to write

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

$\displaystyle\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(2x))dx+\frac{1}{2}\int_{0}^{\pi/2} \ln(1/2)=$

$\displaystyle\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(2x))dx+\frac {\pi\ln(1/2)}{4}$

Now we can do u substitution. u=2x, du=2dx, 1/2du=dx

$\displaystyle\frac{1}{4}\int_{0}^{\pi} \ln(\sin(u))du+\pi/4 \ln(1/2)$

By another use of symmetry

$\displaystyle\int_{0}^{\pi/2} \ln(\sin(u))du=\int_{\pi/2}^{\pi} \ln(\sin(u))du$

$\displaystyle\int_{0}^{\pi} \ln(\sin(u))du=2\int_{0}^{\pi/2} \ln(\sin(u))du$

So we can write

$\displaystyle\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(u))du+\frac {\pi\ln(1/2)}{4}$

$\displaystyle\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(x))dx+\frac {\pi\ln(1/2)}{4}=\int_{0}^{\pi/2} \ln(\sin(x))dx$

$\displaystyle\frac {\pi\ln(1/2)}{4}=\frac{1}{2}\int_{0}^{\pi/2} \ln(\sin(x))dx$

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

$\displaystyle-\frac {\pi\ln(2)}{2}=\int_{0}^{\pi/2} \ln(\sin(x))dx$