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प्रश्न
उत्तर
\[\int_0^\pi \frac{\sin x}{\sin x + \cos x} d x\]
\[ = \frac{1}{2} \int_0^\pi \frac{2\sin x}{\sin x + \cos x} d x\]
\[ = \frac{1}{2} \int_0^\pi \frac{\left( \sin x + \cos x \right) - \left( \cos x - \sin x \right)}{\sin x + \cos x} d x\]
\[ = \frac{1}{2} \int_0^\pi dx - \frac{1}{2} \int_0^\pi \frac{\cos x - \sin x}{\sin x + \cos x}dx\]
\[ = \frac{1}{2} \left[ x \right]_0^\pi - \frac{1}{2} \left[ \log\left| \sin x + \cos x \right| \right]_0^\pi \]
\[ = \frac{1}{2}\left[ \pi - 0 \right] - \frac{1}{2}\left[ \log1 - \log1 \right]\]
\[ = \frac{\pi}{2}\]
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