Advertisements
Advertisements
प्रश्न
उत्तर
\[\text{We have}, \]
\[I = \int \sin^{- 1} \sqrt{x} dx\]
\[\text{ Putting } \sqrt{x} = \sin \theta\]
\[ \Rightarrow x = \sin^2 \theta\]
\[ \Rightarrow dx = 2 \sin \theta \text{ cos } \text{ θ dθ }\]
\[ \Rightarrow dx = \text{ sin}\left( 2\theta \right)d\theta\]
\[ \therefore I = \int \theta \text{ sin } \left( 2\theta \right)d\theta\]
\[ = \theta\left[ \frac{- \text{ cos }2\theta}{2} \right] - \int1\left( \frac{- \text{ cos }2\theta}{2} \right)d\theta\]
\[ = - \frac{\theta \text{ cos }\left( 2\theta \right)}{2} + \frac{1}{2}\int\text{ cos }\left( 2\theta \right)d\theta\]
\[ = - \frac{\theta \text{ cos} \left( 2\theta \right)}{2} + \frac{1}{2}\left[ \frac{\text{ sin} \left( 2\theta \right)}{2} \right] + C\]
\[ = \frac{- \sin^{- 1} \sqrt{x} \left( 1 - 2 \text{ sin}^2 \theta \right)}{2} + \frac{1}{2}\left[ \frac{2 \sin \theta \cos \theta}{2} \right] + C\]
\[ = \frac{- \sin \sqrt{x}\left( 1 - 2x \right)}{2} + \frac{\sin \theta\sqrt{1 - \sin^2 \theta}}{2} + C\]
\[ = \frac{- \sin^{- 1} \sqrt{x} \left( 1 - 2x \right)}{2} + \frac{\sqrt{x} \sqrt{1 - x}}{2} + C\]
\[ = - \frac{1}{2} \sin^{- 1} \left( \sqrt{x} \right) \left( 1 - 2x \right) + \frac{1}{2}\sqrt{x - x^2} + C\]
APPEARS IN
संबंधित प्रश्न
If f' (x) = a sin x + b cos x and f' (0) = 4, f(0) = 3, f
Evaluate the following integrals:
\[\int\frac{x}{\sqrt{8 + x - x^2}} dx\]
\[\int x\ {cosec}^2 \text{ x }\ \text{ dx }\]
\[\int\frac{1}{4 \sin^2 x + 4 \sin x \cos x + 5 \cos^2 x} \text{ dx }\]
