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प्रश्न
Evaluate the following integral:
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उत्तर
\[\text{Let I} = \int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta\]
Consider
\[f\left( - \theta \right) = \log\left( \frac{a - \sin\left( - \theta \right)}{a + \sin\left( - \theta \right)} \right)\]
\[ = \log\left( \frac{a + \sin\theta}{a - \sin\theta} \right) ............\left[ \sin\left( - x \right) = - \sin x \right]\]
\[ = \log \left( \frac{a - \sin\theta}{a + \sin\theta} \right)^{- 1} \]
\[ = - \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right) ..............\left[ \log a^b = b\log a \right]\]
\[ = - f\left( \theta \right)\]
\[\therefore f\left( - \theta \right) = - f\left( \theta \right)\]
\[ \Rightarrow I = \int_{- a}^a \log\left( \frac{a - \sin\theta}{a + \sin\theta} \right)d\theta = 0 .................\left[ \int_{- a}^a f\left( x \right)dx = \begin{cases}2 \int_0^a f\left( x \right)dx, & \text{if }f\left( - x \right) = f\left( x \right) \\ 0, & \text{if }f\left( - x \right) = - f\left( x \right)\end{cases} \right]\]
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