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The integral can be rewritten as

$\displaystyle\int_{0}^{\frac{\pi}{2}} \log(4 + 3\sin(x)) dx - \int_{0}^{\frac{\pi}{2}} \log(4 + 3\cos(x)) dx = L - M.$

For the first of these integrals, namely $L$ , make the substitution $y = \frac{\pi}{2} - x$ . Then $dy = -dx$ and $\sin(x) = \sin(\frac{\pi}{2} - y) = \cos(y)$ . We then have that

$L = \displaystyle\int_{0}^{\frac{\pi}{2}} \log(4 + 3\sin(x)) dx = - \int_{\frac{\pi}{2}}^{0} \log(4 + 3\cos(y)) dy =$

$\displaystyle\int_{0}^{\frac{\pi}{2}} \log(4 + 3\cos(y)) dy.$

Now since $y$ is just a dummy variable at this point we can call it anything we want. So letting it be $x$ , we see that in fact $L = M$ and so the desired solution is $\boxed{0}$ .