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Π ∫ 0 Sin X Sin X + Cos X D X - Mathematics

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प्रश्न

\[\int\limits_0^\pi \frac{\sin x}{\sin x + \cos x} dx\]
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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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अध्याय 20: Definite Integrals - Exercise 20.2 [पृष्ठ ३९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.2 | Q 21 | पृष्ठ ३९

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