Evaluate the integral
\[
\int_{0}^{0.5} \frac{d x}{\sqrt{1-x^{2}}}
\]
Final Answer: \(\boxed{0.5236}\)
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: \(\boxed{0.5236}\)