To evaluate the integral

$\int \frac{\sqrt{4 - x^2}}{x^2} \, dx$

we can use trigonometric substitution. Given that $4 - x^2$ is present in the square root, we’ll use the substitution $x = 2 \sin \theta$, where $dx = 2 \cos \theta \, d\theta$. This substitution will simplify the square root expression.

Step-by-Step Solution

1. Substitute $x = 2 \sin \theta$:

   • Then $dx = 2 \cos \theta \, d\theta$.
   • Substitute $x = 2 \sin \theta$ into the square root term: $\sqrt{4 - x^2} = \sqrt{4 - (2 \sin \theta)^2} = \sqrt{4 - 4 \sin^2 \theta} = \sqrt{4 (1 - \sin^2 \theta)} = \sqrt{4 \cos^2 \theta} = 2 \cos \theta.$

2. Rewrite the integral: Substitute these expressions into the integral: $\int \frac{\sqrt{4 - x^2}}{x^2} \, dx = \int \frac{2 \cos \theta}{(2 \sin \theta)^2} \cdot 2 \cos \theta \, d\theta.$ Simplify the expression: $= \int \frac{2 \cos \theta}{4 \sin^2 \theta} \cdot 2 \cos \theta \, d\theta = \int \frac{4 \cos^2 \theta}{4 \sin^2 \theta} \, d\theta = \int \frac{\cos^2 \theta}{\sin^2 \theta} \, d\theta = \int \cot^2 \theta \, d\theta.$

3. Simplify $\int \cot^2 \theta \, d\theta$: Recall that $\cot^2 \theta = \csc^2 \theta - 1$. So, $\int \cot^2 \theta \, d\theta = \int (\csc^2 \theta - 1) \, d\theta.$ Now integrate each term separately: $= \int \csc^2 \theta \, d\theta - \int 1 \, d\theta = -\cot \theta - \theta + C.$

4. Substitute back in terms of $x$: Recall that $x = 2 \sin \theta$, so $\sin \theta = \frac{x}{2}$. Using the right triangle relationship, we have $\cot \theta = \frac{\sqrt{4 - x^2}}{x}$. Also, from $\sin \theta = \frac{x}{2}$, $\theta = \arcsin \left( \frac{x}{2} \right)$.

   Therefore, the final answer is: $-\frac{\sqrt{4 - x^2}}{x} - \arcsin \left( \frac{x}{2} \right) + C.$

Would you like more details on any of these steps, or have any questions?

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1. How does trigonometric substitution help simplify integrals with radicals?
2. Why do we choose $x = 2 \sin \theta$ specifically for this integral?
3. What other substitution methods could solve similar integrals?
4. How do you interpret the result geometrically?
5. Can trigonometric substitution be applied to functions without square roots?

Tip: When encountering a square root with $a^2 - x^2$, substituting $x = a \sin \theta$ often simplifies the expression.