To solve the given integral using the indicated trigonometric substitution, we will follow these steps:

Given Integral:

$\int \frac{dx}{x^2 \sqrt{4 - x^2}}$ with the substitution $x = 2 \sin \theta$.

Step 1: Trigonometric Substitution

The substitution $x = 2 \sin \theta$ leads to the following:

• $dx = 2 \cos \theta \, d\theta$
• $\sqrt{4 - x^2} = \sqrt{4 - 4\sin^2\theta} = \sqrt{4(1 - \sin^2\theta)} = 2\cos\theta$

Step 2: Substituting into the Integral

Substitute $x = 2\sin\theta$, $dx = 2\cos\theta\,d\theta$, and $\sqrt{4 - x^2} = 2\cos\theta$ into the integral:

$\int \frac{dx}{x^2 \sqrt{4 - x^2}} = \int \frac{2 \cos \theta \, d\theta}{(2 \sin \theta)^2 \cdot 2 \cos \theta}$

This simplifies to: $\int \frac{2 \cos \theta \, d\theta}{4 \sin^2 \theta \cdot 2 \cos \theta} = \int \frac{d\theta}{4 \sin^2 \theta}$ Simplifying further: $\frac{1}{4} \int \csc^2 \theta \, d\theta$

Step 3: Integrating

The integral of $\csc^2 \theta$ is $-\cot \theta$, so we have: $\frac{1}{4} \left( -\cot \theta \right) + C = -\frac{1}{4} \cot \theta + C$

Step 4: Back Substitution

We now need to express $\cot \theta$ in terms of $x$. From the substitution $x = 2\sin\theta$, we know: $\sin \theta = \frac{x}{2}$ Using the Pythagorean identity $\cot \theta = \frac{\cos \theta}{\sin \theta}$, we get: $\cos \theta = \sqrt{1 - \sin^2 \theta} = \sqrt{1 - \frac{x^2}{4}} = \frac{\sqrt{4 - x^2}}{2}$ So: $\cot \theta = \frac{\cos \theta}{\sin \theta} = \frac{\frac{\sqrt{4 - x^2}}{2}}{\frac{x}{2}} = \frac{\sqrt{4 - x^2}}{x}$

Final Answer:

$-\frac{1}{4} \cot \theta + C = -\frac{1}{4} \cdot \frac{\sqrt{4 - x^2}}{x} + C = \frac{\sqrt{4 - x^2}}{4x} + C$

Thus, the evaluated integral is: $\int \frac{dx}{x^2 \sqrt{4 - x^2}} = \frac{\sqrt{4 - x^2}}{4x} + C$

Right Triangle Sketch: