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∫ 2 π 0 Cos − 1 ( Cos X ) D X - Mathematics

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प्रश्न

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]
योग

उत्तर

\[\int_0^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[ = \int_0^\pi \cos^{- 1} \left( \cos x \right)dx +\int_\pi^{2\pi} \cos^{- 1} \left( \cos x \right)dx\]

\[ = \int_0^\pi xdx + \int_\pi^{2\pi} \left( 2\pi - x \right)dx .....................\left[ \pi \leq x \leq 2\pi \Rightarrow - 2\pi \leq - x \leq - \pi \Rightarrow 0 \leq 2\pi - x \leq \pi \right]\]

\[= \left.\frac{x^2}{2}\right|_0^\pi + \left.\frac{\left( 2\pi - x \right)^2}{2 \times \left( - 1 \right)}\right|_\pi^{2\pi} \]
\[ = \frac{1}{2}\left( \pi^2 - 0 \right) - \frac{1}{2}\left( 0 - \pi^2 \right)\]
\[ = \frac{\pi^2}{2} + \frac{\pi^2}{2}\]
\[ = \pi^2\]

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Definite Integrals
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अध्याय 20: Definite Integrals - Exercise 20.3 [पृष्ठ ५६]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 20 Definite Integrals
Exercise 20.3 | Q 28 | पृष्ठ ५६

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