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प्रश्न

\[\int\sqrt{x - x^2} dx\]
बेरीज
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उत्तर

\[\int \sqrt{x - x^2} \text{ dx }\]
\[ = \int \sqrt{- \left( x^2 - x \right)} \text{ dx }\]
\[ = \int \sqrt{- \left\{ x^2 - x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2 \right\}} \text{ dx }\]
\[ = \int \sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2} dx\]
\[ = \left( \frac{x - \frac{1}{2}}{2} \right) \sqrt{x - x^2} + \frac{1}{8} \sin^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right) + C \left[ \because \int\sqrt{a^2 - x^2}dx = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{2x - 1}{4} \right) \sqrt{x - x^2} + \frac{1}{8} \text{ sin}^{- 1} \left( 2x - 1 \right) + C\]

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

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आरडी शर्मा Mathematics [English] Class 12
पाठ 19 Indefinite Integrals
Exercise 19.28 | Q 3 | पृष्ठ १५४

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