Evaluate the integral \[ \int_{0}^{0.5} \frac{d x}{\sqrt{1-x^{2}}} \]

Step 1 ：The problem is to evaluate the definite integral \(\int_{0}^{0.5} \frac{d x}{\sqrt{1-x^{2}}}\).

Step 2 ：The integral is a standard form which is the integral of dx/sqrt(1-x^2) from 0 to 0.5.

Step 3 ：The antiderivative of 1/sqrt(1-x^2) is arcsin(x).

Step 4 ：So, we need to evaluate the antiderivative at the upper limit and lower limit, and then subtract the two results.

Step 5 ：Let's denote the upper limit as b = 0.5 and the lower limit as a = 0.

Step 6 ：After evaluating the antiderivative at the upper limit and lower limit, we get the integral value as approximately 0.5235987755982989.

Step 7 ：So, the integral of the function from 0 to 0.5 is approximately 0.5236.

Step 8 ：Final Answer: \(\boxed{0.5236}\)