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प्रश्न

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

\[\text{ Let I } = \int\sqrt{3 - 2x - 2 x^2}\text{ dx}\]
\[ = \int\sqrt{3 - \left( 2 x^2 + 2x \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x^2 + x \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x^2 + x + \frac{1}{4} - \frac{1}{4} \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x + \frac{1}{2} \right)^2 + \frac{1}{2}}\text{ dx}\]
\[ = \int\sqrt{\frac{7}{2} - 2 \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2}\int\sqrt{\frac{7}{4} - \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2}\int\sqrt{\left( \frac{\sqrt{7}}{2} \right)^2 - \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2} \times \left( \frac{x + \frac{1}{2}}{2} \right) \sqrt{\left( \frac{\sqrt{7}}{2} \right)^2 - \left( x + \frac{1}{2} \right)^2} + \sqrt{2} \times \frac{7}{4 \times 2} \sin^{- 1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}} \right) + C\]
\[ = \frac{2x + 1}{4} \sqrt{3 - 2x - 2 x^2} + \frac{7}{4\sqrt{2}} \sin^{- 1} \left( \frac{2x + 1}{\sqrt{7}} \right) + C\]

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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 11 | पृष्ठ १५४

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