The value of int_0^{pi /2} {log left( {frac{{ | vedclass

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(C) Let $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-

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(C) Let $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right)} \,dx$. Using the property $\int_0^a f(x) \,dx = \int_0^a f(a-x) \,dx$,we get: $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\sin(\pi/2 - x)}}{{4 + 3\cos(\pi/2 - x)}}} \right)} \,dx$. Since $\sin(\pi/2 - x) = \cos x$ and $\cos(\pi/2 - x) = \sin x$,we have: $I = \int_0^{\pi /2} {\log \left( {\frac{{4 + 3\cos x}}{{4 + 3\sin x}}} \right)} \,dx$. Using the property $\log(a/b) = -\log(b/a)$,we get: $I = \int_0^{\pi /2} -\log \left( {\frac{{4 + 3\sin x}}{{4 + 3\cos x}}} \right) \,dx = -I$. Therefore,$2I = 0$,which implies $I = 0$.

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

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