The value of $\displaystyle\int_0^{\pi/2} \log\!\left(\dfrac{4+3\sin x}{4+3\cos x}\right)dx$ is

1. Let $I$ be the integral. Apply $\mathbf{P_4}$ ($x\to\tfrac{\pi}{2}-x$): $\sin$ and $\cos$ swap. 2. So $I=\int_0^{\pi/2}\log\!\left(\dfrac{4+3\cos x}{4+3\sin x}\right)dx=-I$. 3. Hence $2I=0$, giving $I=0$. 4. The swap turns the log argument into its reciprocal, producing $-I$. _Source: syllabus Mathematics Ch 7 "Integrals", p.347_