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प्रश्न
\[\int\limits_{- 1/2}^{1/2} \cos x \log\left( \frac{1 + x}{1 - x} \right) dx\]
उत्तर
\[\int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x\]
\[\text{Let }f(x) = \cos x \log\left( \frac{1 + x}{1 - x} \right)\]
\[\text{Consider }f(- x) = \cos\left( - x \right) \log\left( \frac{1 - x}{1 + x} \right)\]
\[ = \cos x\left\{ - \log\left( \frac{1 + x}{1 - x} \right) \right\} = - \cos x \log\left( \frac{1 + x}{1 - x} \right) = - f\left( x \right)\]
Thus f(x) is an odd function
Therefore,
\[ \int_\frac{- 1}{2}^\frac{1}{2} \cos x \log\left( \frac{1 + x}{1 - x} \right) d x = 0\]
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