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∫ 3 π 2 π √ 1 − Cos 2 X D X - Mathematics

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प्रश्न

\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]
योग
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उत्तर

\[\int_\pi^\frac{3\pi}{2} \sqrt{1 - \cos2x}dx\]
\[= \int_\pi^\frac{3\pi}{2} \sqrt{2 \sin^2 x}dx\]
\[ = \sqrt{2} \int_\pi^\frac{3\pi}{2} \left| \sin x \right|dx\]
\[ = - \sqrt{2} \int_\pi^\frac{3\pi}{2} \sin x\ dx .................\left( \sin x < 0 for\ \pi \leq x \leq 2\pi \right)\]

\[= - \sqrt{2}\left( - \cos x \right) |_\pi^\frac{3\pi}{2} \]
\[ = \sqrt{2}\left( \cos\frac{3\pi}{2} - cos\pi \right)\]
\[ = \sqrt{2} \left[ 0 - \left( - 1 \right) \right]\]
\[ = \sqrt{2} \times 1\]
\[ = \sqrt{2}\]

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अध्याय 20: Definite Integrals - Exercise 20.1 [पृष्ठ १८]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.1 | Q 62 | पृष्ठ १८

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