$\begin{aligned} & \text {} \int \frac{2 \sin 2 x-3 \cos x}{2 \sin ^2 x-3 \sin x+4} d x=\int \frac{\cos x(4 \sin x-3)}{2 \sin ^2 x-3 \sin x+4} d x \\ & \text { Let } \sin x=t, \cos x d x=d t=\int \frac{(4 t-3)}{2 t^2-3 t+4} d t \\ & 2 t^2-3 t+4=u \Rightarrow(4 t-3) d t=d u \\ & \therefore \quad \int \frac{d u}{u}=\ln |u|+C=\ln \left|2 t^2-3 t+4\right|+C \\ & \quad=\ln \left(2 \sin ^2 x-3 \sin x+4\right)+C \\ & \therefore \quad f(x)=\ln \left(2 \sin ^2 x-3 \sin x+4\right) \\ & f\left(\frac{\pi}{2}\right)-f(0)=\ln 3-\ln 4=\ln \left(\frac{3}{4}\right)=\log \left(\frac{3}{4}\right) .\end{aligned}$