Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{ Let I } = \int\sqrt{3 - 2x - 2 x^2}\text{ dx}\]
\[ = \int\sqrt{3 - \left( 2 x^2 + 2x \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x^2 + x \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x^2 + x + \frac{1}{4} - \frac{1}{4} \right)}\text{ dx}\]
\[ = \int\sqrt{3 - 2 \left( x + \frac{1}{2} \right)^2 + \frac{1}{2}}\text{ dx}\]
\[ = \int\sqrt{\frac{7}{2} - 2 \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2}\int\sqrt{\frac{7}{4} - \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2}\int\sqrt{\left( \frac{\sqrt{7}}{2} \right)^2 - \left( x + \frac{1}{2} \right)^2}\text{ dx}\]
\[ = \sqrt{2} \times \left( \frac{x + \frac{1}{2}}{2} \right) \sqrt{\left( \frac{\sqrt{7}}{2} \right)^2 - \left( x + \frac{1}{2} \right)^2} + \sqrt{2} \times \frac{7}{4 \times 2} \sin^{- 1} \left( \frac{x + \frac{1}{2}}{\frac{\sqrt{7}}{2}} \right) + C\]
\[ = \frac{2x + 1}{4} \sqrt{3 - 2x - 2 x^2} + \frac{7}{4\sqrt{2}} \sin^{- 1} \left( \frac{2x + 1}{\sqrt{7}} \right) + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = x + b, f(1) = 5, f(2) = 13, find f(x)
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
If \[\int\frac{1}{\left( x + 2 \right)\left( x^2 + 1 \right)}dx = a\log\left| 1 + x^2 \right| + b \tan^{- 1} x + \frac{1}{5}\log\left| x + 2 \right| + C,\] then
\[\int\frac{1 + \sin x}{\sin x \left( 1 + \cos x \right)} \text{ dx }\]
\[ \int\left( 1 + x^2 \right) \ \cos 2x \ dx\]
