Question

integrate: x^2/sqrt (1-x^2)

Answer 1

$\frac{1}{2}\arcsin x - \frac{x\sqrt{1-x^2}}{2} + C$

Answer 2

To integrate the given function $\frac{x^2}{\sqrt{1-x^2}}$, we use trigonometric substitution.

1. Choose the appropriate substitution: The integrand contains the term $\sqrt{1-x^2}$, which suggests the substitution $x = \sin \theta$. From this, we have: $dx = \cos \theta \, d\theta$ And $\sqrt{1-x^2} = \sqrt{1-\sin^2 \theta} = \sqrt{\cos^2 \theta} = |\cos \theta|$. Assuming the principal value branch for $\theta$ (i.e., $-\frac{\pi}{2} \le \theta \le \frac{\pi}{2}$), $\cos \theta \ge 0$, so $\sqrt{1-x^2} = \cos \theta$.

2. Substitute into the integral: $\int \frac{x^2}{\sqrt{1-x^2}} dx = \int \frac{(\sin \theta)^2}{\cos \theta} (\cos \theta \, d\theta)$ $= \int \sin^2 \theta \, d\theta$

3. Use trigonometric identity to simplify the integrand: We use the power-reduction formula for $\sin^2 \theta$: $\sin^2 \theta = \frac{1 - \cos(2\theta)}{2}$ Substitute this into the integral: $\int \frac{1 - \cos(2\theta)}{2} \, d\theta$ $= \frac{1}{2} \int (1 - \cos(2\theta)) \, d\theta$

4. Integrate with respect to $\theta$: $= \frac{1}{2} \left( \int 1 \, d\theta - \int \cos(2\theta) \, d\theta \right)$ $= \frac{1}{2} \left( \theta - \frac{\sin(2\theta)}{2} \right) + C$ $= \frac{\theta}{2} - \frac{\sin(2\theta)}{4} + C$

5. Convert the result back to $x$: We have $x = \sin \theta$, so $\theta = \arcsin x$. For $\sin(2\theta)$, we use the double angle identity: $\sin(2\theta) = 2 \sin \theta \cos \theta$. We know $\sin \theta = x$. And $\cos \theta = \sqrt{1-\sin^2 \theta} = \sqrt{1-x^2}$ (since $\cos \theta \ge 0$ for the chosen range of $\theta$). So, $\sin(2\theta) = 2x\sqrt{1-x^2}$.

Substitute these back into the expression: $\frac{\arcsin x}{2} - \frac{2x\sqrt{1-x^2}}{4} + C$ $= \frac{\arcsin x}{2} - \frac{x\sqrt{1-x^2}}{2} + C$

The final answer is $\frac{1}{2}\arcsin x - \frac{x\sqrt{1-x^2}}{2} + C$.

Explanation of the solution: The integral $\int \frac{x^2}{\sqrt{1-x^2}} dx$ is solved using the trigonometric substitution $x = \sin \theta$. This transforms the integral into $\int \sin^2 \theta \, d\theta$. Using the identity $\sin^2 \theta = \frac{1-\cos(2\theta)}{2}$, the integral becomes $\frac{1}{2}\int (1-\cos(2\theta)) \, d\theta$. Integrating this yields $\frac{1}{2}(\theta - \frac{\sin(2\theta)}{2}) + C$. Finally, substituting back $\theta = \arcsin x$ and $\sin(2\theta) = 2\sin\theta\cos\theta = 2x\sqrt{1-x^2}$ gives the result $\frac{1}{2}\arcsin x - \frac{x\sqrt{1-x^2}}{2} + C$.