A common integral in disguise

$\displaystyle\int_0^{\pi/2} \ln (\cos( x))dx =I=\displaystyle\int_0^{\pi/2} \ln (\sin (x)) dx.$

By symmetry we have $\ln (\cos (x))=\ln (\sin (x))$ in $[0,\pi/2]$ ,Thus we have:

$2I=\displaystyle\int_0^{\pi/2}\ln (\cos (x)) dx+ \displaystyle\int_0^{\pi/2} \ln (\sin (x)) dx= \displaystyle\int_0^{\pi/2} \ln(\sin (x) \cos (x))dx=\displaystyle\int_0^{\pi/2} \ln\big(\frac{1}{2}\cdot\sin(2x)\big) dx.$

( I used $\ln(a\cdot b)=\ln(a)+\ln(b)$ )

Now we split the integral back up to obtain

$-\displaystyle\int_0^{\pi/2}\ln(2)dx+\displaystyle\int_0^{\pi/2}\ln(\sin(2x))dx=2I.$

Thus we can now substitute $u=2x$ to obtain

$-\dfrac{\pi\ln(2)}{2}+\dfrac{1}{2}\displaystyle\int_0^\pi \ln (\sin (u)) du=2I$

But the integral of $\ln (\sin (u))$ is $2I$ , thus we have

$-\dfrac{\pi\ln(2)}{2}+I=2I, \ \to {\boxed{I=-\dfrac{\pi \ln(2)}{2}.}}$