Integration and domain Let have for example 
$$f(x) = \frac{\sin^2 x}{1- \cos x}$$ 
And let say that we want calculate:
$$ \int_{-\pi/2}^{\pi/2} f(x) \, dx $$
After transform: 
$$f(x) = \frac{\sin^2 x}{1- \cos x} = 1 + \cos x$$
so
$$ \int_{-\pi/2}^{\pi/2} f(x) \, dx  = x + \sin x + C := F(x)$$
But $f$ was not defined on $x = 0$. So if we interpret this integral as space under graph then we have hole at $x=0$. I know that
$$\int_{-\pi/2}^{\pi/2} f(x) = F(\pi/2) - F(-\pi/2) $$
but why we can ignore lack of domain at $x=0$?
 A: We don't ignore it. While integrating, the actual integration is  from $-\pi/2 \to 0^-$ and $0^+ \to \pi/2$ and the sum of both the integrals.
A: You can define the function at $0$ to anything, the integral won't change. The measure of the set $\{0\}$ is zero, so the value doesn't affect the integral at all. Moreover, you can redefine the function at a set of zero measure to anything else and the integral wouldn't change.
A: For your particular function, $x = 0$ is a removable singularity. That is, we can define an extension $F(x)$ which is equal to $f(x)$ on $[-\pi/2,\pi/2]-\{0\}$ and some other value at $0$. As mentioned in the other answers, the integrals of functions that differ on a set of measure zero have the same values over the same domain (presuming the integrals make sense on that domain), so $\int_{-\pi/2}^{\pi/2}f(x)dx = \int_{-\pi/2}^{\pi/2}F(x)dx$. 
What is NOT true is that an integral like 
$$
\int_{-1}^{1}\frac{dx}{x^2}
$$ 
could be handled with the same trick, since in this case the singularity in $1/x^2$ at $x = 0$ is not removable. 
