Advertisements
Advertisements
प्रश्न
\[\int\sqrt{1 + 2x - 3 x^2}\text{ dx } \]
बेरीज
उत्तर
\[\int\sqrt{1 + 2x - 3 x^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{2}{3}x - x^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left( x^2 - \frac{2}{3}x \right)}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left\{ x^2 - \frac{2}{3}x + \left( \frac{1}{3} \right)^2 - \left( \frac{1}{3} \right)^2 \right\}}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{1}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{4}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3} \left[ \frac{\left( x - \frac{1}{3} \right)}{2} \sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2} + \frac{\left( \frac{2}{3} \right)^2}{2} \sin^{- 1} \left( \frac{x - \frac{1}{3}}{\frac{2}{3}} \right) \right] + C ................................\left[ \because \int\sqrt{a^2 - x^2}dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{3x - 1}{6} \right) \sqrt{1 + 2x - 3 x^2} + \frac{2\sqrt{3}}{9} \sin^{- 1} \left( \frac{3x - 1}{2} \right) + C\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{2}{3}x - x^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left( x^2 - \frac{2}{3}x \right)}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} - \left\{ x^2 - \frac{2}{3}x + \left( \frac{1}{3} \right)^2 - \left( \frac{1}{3} \right)^2 \right\}}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{1}{3} + \frac{1}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\frac{4}{9} - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3}\int\sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2}dx\]
\[ = \sqrt{3} \left[ \frac{\left( x - \frac{1}{3} \right)}{2} \sqrt{\left( \frac{2}{3} \right)^2 - \left( x - \frac{1}{3} \right)^2} + \frac{\left( \frac{2}{3} \right)^2}{2} \sin^{- 1} \left( \frac{x - \frac{1}{3}}{\frac{2}{3}} \right) \right] + C ................................\left[ \because \int\sqrt{a^2 - x^2}dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{3x - 1}{6} \right) \sqrt{1 + 2x - 3 x^2} + \frac{2\sqrt{3}}{9} \sin^{- 1} \left( \frac{3x - 1}{2} \right) + C\]
shaalaa.com
या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
APPEARS IN
संबंधित प्रश्न
\[\int\frac{\left( 1 + x \right)^3}{\sqrt{x}} dx\]
\[\int\frac{2 x^4 + 7 x^3 + 6 x^2}{x^2 + 2x} dx\]
\[\int \cos^{- 1} \left( \sin x \right) dx\]
\[\int\frac{1}{1 - \sin\frac{x}{2}} dx\]
\[\int\frac{1}{\text{cos}^2\text{ x }\left( 1 - \text{tan x} \right)^2} dx\]
\[\int \sin^2\text{ b x dx}\]
\[\int\frac{- \sin x + 2 \cos x}{2 \sin x + \cos x} dx\]
\[\int\frac{\sin 2x}{\left( a + b \cos 2x \right)^2} dx\]
\[\int\frac{x^5}{\sqrt{1 + x^3}} dx\]
` ∫ tan^5 x sec ^4 x dx `
` ∫ sec^6 x tan x dx `
\[\int \cot^6 x \text{ dx }\]
\[\int\frac{1}{\sin x \cos^3 x} dx\]
Evaluate the following integrals:
\[\int\frac{x^7}{\left( a^2 - x^2 \right)^5}dx\]
\[\int\frac{x}{x^4 + 2 x^2 + 3} dx\]
\[\int\frac{\cos x - \sin x}{\sqrt{8 - \sin2x}}dx\]
\[\int\frac{x + 1}{x^2 + x + 3} dx\]
\[\int\frac{x + 1}{\sqrt{4 + 5x - x^2}} \text{ dx }\]
\[\int\frac{1}{4 \cos^2 x + 9 \sin^2 x}\text{ dx }\]
\[\int\frac{1}{1 + 3 \sin^2 x} \text{ dx }\]
\[\int\frac{1}{3 + 2 \cos^2 x} \text{ dx }\]
\[\int\frac{1}{\left( \sin x - 2 \cos x \right)\left( 2 \sin x + \cos x \right)} \text{ dx }\]
\[\int\frac{2 \tan x + 3}{3 \tan x + 4} \text{ dx }\]
\[\int x^2 \text{ cos x dx }\]
\[\int\frac{\log \left( \log x \right)}{x} dx\]
\[\int \cos^{- 1} \left( 4 x^3 - 3x \right) \text{ dx }\]
\[\int\frac{x^3 \sin^{- 1} x^2}{\sqrt{1 - x^4}} \text{ dx }\]
\[\int\frac{3 + 4x - x^2}{\left( x + 2 \right) \left( x - 1 \right)} dx\]
\[\int\frac{x^2 + x + 1}{\left( x + 1 \right)^2 \left( x + 2 \right)} dx\]
\[\int\frac{1}{x^4 + x^2 + 1} \text{ dx }\]
\[\int\frac{x}{\left( x - 3 \right) \sqrt{x + 1}} dx\]
\[\int\frac{x}{\left( x^2 + 2x + 2 \right) \sqrt{x + 1}} \text{ dx}\]
\[\int\frac{x}{\left( x^2 + 4 \right) \sqrt{x^2 + 1}} \text{ dx }\]
\[\int\frac{1}{\cos x + \sqrt{3} \sin x} \text{ dx } \] is equal to
\[\int\frac{\sin x}{1 + \sin x} \text{ dx }\]
\[\int\frac{1}{1 - x - 4 x^2}\text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
\[\int \tan^{- 1} \sqrt{x}\ dx\]
\[\int\frac{1}{\left( x^2 + 2 \right) \left( x^2 + 5 \right)} \text{ dx}\]
\[\int\frac{\cot x + \cot^3 x}{1 + \cot^3 x} \text{ dx}\]