Let $I = \int_{0}^{\frac{\pi}{2}}[\log(\sin~x)-\log(2\cos~x)]dx$

$I = \int_{0}^{\frac{\pi}{2}}\log(\sin~x)dx - \int_{0}^{\frac{\pi}{2}}\log(2\cos~x)dx$

$I = \int_{0}^{\frac{\pi}{2}}\log(\sin~x)dx - \int_{0}^{\frac{\pi}{2}}[\log 2 + \log(\cos~x)]dx$

$I = \int_{0}^{\frac{\pi}{2}}\log(\sin~x)dx - \int_{0}^{\frac{\pi}{2}}\log 2~dx - \int_{0}^{\frac{\pi}{2}}\log(\cos~x)dx$

We know that $\int_{0}^{\frac{\pi}{2}}\log(\sin~x)dx = \int_{0}^{\frac{\pi}{2}}\log(\cos~x)dx = -\frac{\pi}{2}\log 2$

Therefore, $I = -\frac{\pi}{2}\log 2 - \log 2 \int_{0}^{\frac{\pi}{2}}dx - (-\frac{\pi}{2}\log 2)$

$I = -\frac{\pi}{2}\log 2 - \log 2 [x]_{0}^{\frac{\pi}{2}} + \frac{\pi}{2}\log 2$

$I = -\frac{\pi}{2}\log 2 - \log 2 (\frac{\pi}{2} - 0) + \frac{\pi}{2}\log 2$

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

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