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The given integral is:

\[
\int_0^{1/2} \frac{6}{\sqrt{1 - x^2}} \, dx
\]

This resembles the standard integral for \(\arcsin(x)\), since the derivative of \(\arcsin(x)\) is:

\[
\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}
\]

Thus, we can solve this integral using substitution with \(\arcsin(x)\):

\[
\int \frac{1}{\sqrt{1 - x^2}} \, dx = \arcsin(x) + C
\]

In this case, we have an extra constant factor of 6. So, the integral becomes:

\[
6 \int_0^{1/2} \frac{1}{\sqrt{1 - x^2}} \, dx = 6 [\arcsin(x)]_0^{1/2}
\]

Now, we compute the definite integral:

\[
= 6 [\arcsin(1/2) - \arcsin(0)]
\]

We know that:

\[
\arcsin(1/2) = \frac{\pi}{6} \quad \text{and} \quad \arcsin(0) = 0
\]

Therefore, the solution is:

\[
6 \times \left(\frac{\pi}{6} - 0\right) = \pi
\]

Thus, the value of the integral is:

\[
\boxed{\pi}
\]

Would you like further details on any of the steps, or have any questions?

Here are 5 related questions for further exploration:
1. How does the derivative of \(\arcsin(x)\) relate to this integral?
2. What are other standard integrals that involve inverse trigonometric functions?
3. How can the limits of integration be changed to affect the result?
4. What happens if we introduce different constants in front of the integral?
5. Can this integral be solved using another method, like trigonometric substitution?

**Tip**: When dealing with integrals involving \(\sqrt{1 - x^2}\), remember that inverse trigonometric functions like \(\arcsin(x)\) often come into play.

Evaluate the integral: \int_0^{1/2} \frac{6}{\sqrt{1 - x^2}} dx.

This integral involves inverse trigonometric functions and is solved using the arcsin formula. Learn how to evaluate \int_0^{1/2} \frac{6}{\sqrt{1 - x^2}} dx, suitable for undergraduate students studying calculus.

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