Question

\[\int\sqrt{1 + x - 2 x^2} \text{ dx }\]

Answer 1

\[\int \sqrt{1 + x - 2 x^2} \text{ dx}\]
\[ = \int \sqrt{2\left( \frac{1}{2} + \frac{x}{2} - x^2 \right)} \text{ dx}\]
\[ = \sqrt{2} \int\sqrt{\frac{1}{2} - \left( x^2 - \frac{x}{2} \right)} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{\frac{1}{2} - \left( x^2 - \frac{x}{2} + \frac{1}{4^2} - \frac{1}{4^2} \right)} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{\frac{1}{2} + \frac{1}{16} - \left( x - \frac{1}{4} \right)^2} \text{ dx}\]
\[ = \sqrt{2} \int \sqrt{\left( \frac{3}{4} \right)^2 - \left( x - \frac{1}{4} \right)^2} \text{ dx}\]
\[ = \sqrt{2} \left[ \left( \frac{x - \frac{1}{4}}{2} \right) \sqrt{\left( \frac{3}{4} \right)^2 - \left( x - \frac{1}{4} \right)^2} + \frac{9}{32} \sin^{- 1} \left( \frac{x - \frac{1}{4}}{\frac{3}{4}} \right) \right] + C \left[ \because \int\sqrt{a^2 - x^2}\text{ dx} = \frac{1}{2}x\sqrt{a^2 - x^2} + \frac{1}{2} a^2 \sin^{- 1} \frac{x}{a} + C \right]\]
\[ = \left( \frac{4x - 1}{8} \right) \sqrt{1 + x - 2 x^2} + \frac{9\sqrt{2}}{32} \sin^{- 1} \left( \frac{4x - 1}{3} \right) + C\]