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प्रश्न
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उत्तर
\[Let\ I = \int_{- \frac{\pi}{2}}^\frac{\pi}{2} \log\left( \frac{2 - \sin x}{2 + \sin x} \right) d x\]
\[Here, f\left( x \right) = log\left( \frac{2 - \sin x}{2 + \sin x} \right)\]
\[f\left( - x \right) = log\left( \frac{2 - \sin\left( - x \right)}{2 + \sin\left( - x \right)} \right) = log\left( \frac{2 + \sin x}{2 - \sin x} \right) = - log\left( \frac{2 - \sin x}{2 + \sin x} \right) = - f\left( x \right)\]
\[\text{Hence} f\left( x \right) \text{is an odd function}\]
\[ \therefore I = 0\]
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