I would take the whole area and subtract the non shaded parts
$\displaystyle \displaystyle \int^{\pi /2}_0 \cos(x) \,dx - \left( \int^{\pi /6}_0 \cos(x) \,dx + \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{\pi}{2}\right)$
in case you're wondering the last term is the area of a triangle
Don't quote me on it though, someone may have a better answer!
You could also calculate
$\displaystyle \displaystyle\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}c osxdx+\frac{1}{2}\frac{\pi}{6}\frac{\sqrt{3}}{2}-\frac{1}{2}\frac{1}{2}\frac{\pi}{3}$
In other words, the area under the curve from $\displaystyle \frac{\pi}{6}\rightarrow\frac{\pi}{3}$
then add in the area of the upper triangle and subtract the lower one..
@ e^(i*pi)
the last part isn't a triangle (the curve is cosx there).