Knowing that the function
$$\frac{1}{\sqrt{1-x^2}}\ $$ is defined only for $-1 < x <1$, are the limits of integration below allowed?
$$\int_{-1}^1 \frac{1}{\sqrt{1-x^2}} \,dx= \pi$$
PS: As stated above, I can calculate the result as $\pi$, I am just questioning the reasoning behind using $-1$ and $1$ as limits of integration when the function isn't defined at these points.
This is an improper integral. Interpret it as follows:
$$\int_{-1}^1 \frac{1}{\sqrt {1-x^2}} \,dx= \lim_{m \to -1^+} \lim_{n \to 1^-} \int_{m}^n \frac{1}{\sqrt {1-x^2}} \,dx = \lim_{m \to -1^+} \lim_{n \to 1^-} \left[ \arcsin x \right]_{m}^n = \pi.$$