Question 5 (10 points) Use a trig. substitution to evaluate \( \int \frac{\sqrt{4-4 x^{2}}}{x^{2}} d x \)

To evaluate the integral \( \int \frac{\sqrt{4-4 x^{2}}}{x^{2}} d x \) using trigonometric substitution, we can start by simplifying the radical expression. Notice that \( \sqrt{4 - 4x^{2}} = 2\sqrt{1 - x^{2}} \). So the integral simplifies to: \[ \int \frac{2\sqrt{1 - x^{2}}}{x^{2}} d x = 2 \int \frac{\sqrt{1 - x^{2}}}{x^{2}} d x \] Next, we use the substitution \( x = 2 \sin(\theta) \), which gives \( dx = 2 \cos(\theta) d\theta \). The limits for \( x \) will change accordingly if we were to calculate a definite integral, but we will keep them more general here. Under this substitution, we also have: \[ \sqrt{1 - x^{2}} = \sqrt{1 - (2\sin(\theta))^{2}} = \sqrt{1 - 4\sin^2(\theta)} = 2\cos(\theta) \sqrt{1 - \sin^2(\theta)} = 2\cos^2(\theta) \] Now substituting all of this into our integral, we have: \[ 2 \int \frac{2\cos^2(\theta)}{(2\sin(\theta))^{2}} \cdot 2\cos(\theta) d\theta = 2 \int \frac{2\cos^2(\theta)}{4\sin^2(\theta)} \cdot 2\cos(\theta) d\theta \] \[ = 2 \int \frac{4\cos^3(\theta)}{4\sin^2(\theta)} d\theta = 2 \int \frac{\cos^3(\theta)}{\sin^2(\theta)} d\theta \] Now, this integral can be expressed as: \[ = 2 \int \cos^3(\theta) \csc^2(\theta) d\theta \] Using the identity \( \csc^2(\theta) = 1 + \cot^2(\theta) \), we can rewrite the integral as \( 2 \int \cos^3(\theta) \left(1 + \cot^2(\theta)\right) d\theta \). This leads to the expression that allows further integration by parts or more familiar techniques based on the functions available. Once this integral is solved, we substitute back to find the answer in terms of \( x \). This procedure ultimately leads to the result of the integral, after managing all transformations and substitutions accordingly.