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∫ √ 3 + 2 X − X 2 D X - Mathematics

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प्रश्न

\[\int\sqrt{3 + 2x - x^2} \text{ dx}\]
योग
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उत्तर

\[\int \sqrt{3 + 2x - x^2} \text{ dx}\]
\[ = \int \sqrt{3 - \left( x^2 - 2x \right)}\text{ dx}\]
\[ = \int \sqrt{3 - \left( x^2 - 2x + 1 - 1 \right)}\text{ dx}\]
\[ = \int \sqrt{4 - \left( x - 1 \right)^2}\text{ dx}\]
\[ = \int \sqrt{2^2 - \left( x - 1 \right)^2} \text{ dx} \left[ \because \int\sqrt{a^2 - x^2}\text{ dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \text{ sin }^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{x - 1}{2} \right) \sqrt{2^2 - \left( x - 1 \right)^2} + \frac{2^2}{2} \sin^{- 1} \left( \frac{x - 1}{2} \right) + C\]
\[ = \frac{x - 1}{2}\sqrt{3 + 2x - x^2} + \sin^{- 1} \left( \frac{x - 1}{2} \right) + C\]

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अध्याय 19: Indefinite Integrals - Exercise 19.28 [पृष्ठ १५४]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 19 Indefinite Integrals
Exercise 19.28 | Q 1 | पृष्ठ १५४

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