Question
$\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \sin x}{4+3 \cos x}\right) d x$ का मान है:
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}}\left[\frac{4+3 \sin \left(\frac{\pi}{2}-x\right)}{4+3 \cos \left(\frac{\pi}{2}-x\right)}\right] d x$ $\left(\int_{0}^{a} f(x) d x=\int_{0}^{a} f(a-x) d x\right)$
$\Rightarrow I=\int_{0}^{\frac{\pi}{2}} \log \left(\frac{4+3 \cos x}{4+3 \sin x}\right) d x$.....$(2)$
Adding $( 1 )$ and $( 2 )$, we obtain
$2 I=\int_{0}^{\frac{\pi}{2}}\left\{\log \left(\frac{4+3 \sin x}{4+3 \cos x}\right)+\log \left(\frac{4+3 \cos x}{4+3 \sin x}\right)\right\} d x$
$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}}\left(\frac{4+3 \sin x}{4+3 \cos x} \times \frac{4+3 \cos x}{4+3 \sin x}\right) d x$
$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} \log 1 d x$
$\Rightarrow 2 I=\int_{0}^{\frac{\pi}{2}} 0 d x$
$\Rightarrow I=0$
Hence, the correct Answer is $B$
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