int_0^{pi / 2} log left(frac{4+3 sin x}{4+3 c | vedclass

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(A) Let $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ ... $(1)$Using the property $\int_0^a f(x) dx = \int_0^a f(a-x) dx$

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(A) Let $I = \int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}ight) d x$ ... $(1)$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$$I = \int_0^{\pi / 2} \log \left(\frac{4+3 \cos x}{4+3 \sin x}ight) d x$ ... $(2)$Adding $(1)$ and $(2)$:$2I = \int_0^{\pi / 2} \left[ \log \left(\frac{4+3 \sin x}{4+3 \cos x}ight) + \log \left(\frac{4+3 \cos x}{4+3 \sin x}ight) ight] d x$$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$Since $\log(1) = 0$,we have $2I = 0$,which implies $I = 0$.

$\int_0^{\pi / 2} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x=$

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