The value of the integral int_0^{pi / 2} log | vedclass

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

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$0$

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(C) Let $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}ight) d x$. Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$,we get: $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin(\pi/2 - x)}{4+3 \cos(\pi/2 - x)}ight) d x$ $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}ight) d x$. Adding the two expressions for $I$: $2I = \int_0^{\pi / 2} \left[ \log \left(\frac{4+3 \sin x}{4+3 \cos x}ight) + \log \left(\frac{4+3 \cos x}{4+3 \sin x}ight) ight] d x$ $2I = \int_0^{\pi / 2} \log \left( \frac{4+3 \sin x}{4+3 \cos x} 	imes \frac{4+3 \cos x}{4+3 \sin x} ight) d x$ $2I = \int_0^{\pi / 2} \log(1) d x = \int_0^{\pi / 2} 0 d x = 0$. Therefore,$I = 0$.

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

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