QuestionEvaluate the integral: $\int_{0}^{0.4} \frac{d x}{\sqrt{1-x^{2}}}$

Studdy Solution
Compute the values of the inverse sine function:
1. $\sin^{-1}(0.4)$ is the angle whose sine is 0.4.
2. $\sin^{-1}(0)$ is the angle whose sine is 0, which is 0.
Thus, the definite integral is:
$$\sin^{-1}(0.4) - 0 = \sin^{-1}(0.4)$$
The value of the integral is:
$$\boxed{\sin^{-1}(0.4)}$$