We are tasked with evaluating the integral:

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

and showing that the substitution $x = 2 \sin(\theta)$ transforms this integral into $\int d\theta$, then evaluating $I$ in terms of $\theta$.

Step 1: Substitution

Given $x = 2 \sin(\theta)$, we differentiate this to find $dx$:

$dx = 2 \cos(\theta) d\theta$

Now, we substitute $x = 2 \sin(\theta)$ into the square root term $\sqrt{4 - x^2}$:

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

Step 2: Substituting into the Integral

Now, substitute both $dx = 2 \cos(\theta) d\theta$ and $\sqrt{4 - x^2} = 2 \cos(\theta)$ into the original integral:

$I = \int \frac{2 \cos(\theta) d\theta}{2 \cos(\theta)}$

The $2 \cos(\theta)$ terms cancel out, leaving:

$I = \int d\theta$

Step 3: Evaluating the Integral

The integral of $d\theta$ is simply:

$I = \theta + C$

Step 4: Express $\theta$ in Terms of $x$

Since $x = 2 \sin(\theta)$, we have:

$\sin(\theta) = \frac{x}{2}$

Thus, $\theta = \arcsin\left(\frac{x}{2}\right)$. Therefore, the final answer for $I$ in terms of $x$ is:

$I = \arcsin\left(\frac{x}{2}\right) + C$

Would you like further details or have any questions?

Here are five related questions to expand on this:

1. How do we apply trigonometric substitutions for other integrals involving square roots?
2. What is the general formula for integrating $\frac{dx}{\sqrt{a^2 - x^2}}$?
3. How does this method relate to the inverse trigonometric functions?
4. Can the same substitution be used for definite integrals? How?
5. What are some alternative substitution methods for this integral?

Tip: Trigonometric substitutions are especially useful when dealing with integrals involving square roots of quadratic expressions.