मराठी

∫ √ 9 − X 2 D X

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प्रश्न

\[\int\sqrt{9 - x^2}\text{ dx}\]
बेरीज
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उत्तर

\[\int \sqrt{9 - x^2} \text{ dx}\]
\[ = \int \sqrt{3^2 - x^2} \text{ dx}\]
\[ = \frac{x}{2}\sqrt{3^2 - x^2} + \frac{3^2}{2} \sin^{- 1} \left( \frac{x}{3} \right) + C \left[ \because \int\sqrt{a^2 - x^2} \text{ dx } = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \text{ sin}^{- 1} \left( \frac{x}{a} \right) + C \right]\]
\[ = \frac{x}{2}\sqrt{9 - x^2} + \frac{9}{2} \sin^{- 1} \left( \frac{x}{3} \right) + C\]

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पाठ 18: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

APPEARS IN

आर.डी. शर्मा Mathematics Volume 1 and 2 [English] Class 12
पाठ 18 Indefinite Integrals
Exercise 19.28 | Q 7 | पृष्ठ १५४

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