To evaluate the given integral, the formula associated with arcsine (arcsin) is likely required, as the integrand resembles the derivative of $\arcsin(u)$. The integral can be analyzed step-by-step as follows:

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Integral Setup:

$\int_0^{1/4} \frac{1}{\sqrt{1 - (2x)^2}} \, dx$

Step 1: Recognize the substitution

This form suggests the use of the arcsin integral formula: $\int \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(u) + C$

Here, $u = 2x$. To simplify:

1. Substitute $u = 2x$, so $du = 2 \, dx$.
2. Therefore, $dx = \frac{du}{2}$.

Change the limits accordingly:

• When $x = 0$, $u = 0$.
• When $x = \frac{1}{4}$, $u = \frac{1}{2}$.

Step 2: Update the integral

$\int_0^{1/4} \frac{1}{\sqrt{1 - (2x)^2}} \, dx = \frac{1}{2} \int_0^{1/2} \frac{1}{\sqrt{1 - u^2}} \, du$

Step 3: Evaluate the integral

Using the arcsin formula:

$\int \frac{1}{\sqrt{1 - u^2}} \, du = \arcsin(u)$

Apply this result: $\frac{1}{2} \int_0^{1/2} \frac{1}{\sqrt{1 - u^2}} \, du = \frac{1}{2} \big[\arcsin(u)\big]_0^{1/2}$

Step 4: Compute the definite integral

[ \frac{1}{2} \big[\arcsin(u)\big]_0^{1/2} = \frac{1}{2} \left[\arcsin\left(\frac{1}{2}\right) - \arcsin(0)\right] ]

From trigonometric values:

• $\arcsin\left(\frac{1}{2}\right) = \frac{\pi}{6}$,
• $\arcsin(0) = 0$.

Thus: [ \frac{1}{2} \left[\arcsin\left(\frac{1}{2}\right) - \arcsin(0)\right] = \frac{1}{2} \left[\frac{\pi}{6} - 0\right] = \frac{\pi}{12}. ]

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Final Answer:

$\int_0^{1/4} \frac{1}{\sqrt{1 - (2x)^2}} \, dx = \frac{\pi}{12}.$

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Let me know if you'd like a more detailed breakdown! Here are some related questions for practice:

1. What is the formula for the derivative of $\arcsin(x)$?
2. How would the integral change if the limits were different, say $0$ to $1/2$?
3. What is the value of $\arcsin(1)$ and why?
4. Could we solve this problem using trigonometric substitution? If so, how?
5. How is the arcsin function connected to right-angle triangles?

Tip: Always confirm the domain of your integrand to ensure the solution is valid, especially when working with functions like $\arcsin(x)$ and $\arccos(x)$.