Does the integral $\int_{a}^{b}\frac{dx}{\sqrt{(x-a)(x-b)}}$ exist?

We first show that the integral is independent of $a$ and $b$, and then let $a=-1$ and $b=1$ to get the result that $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\pi$$

For the first part, use the substitution $x=a+t(b-a)$ $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}=\int_0^1\frac{(b-a)dt}{\sqrt{t(b-a)(1-t)(b-a)}}=\int_0^1\frac{dt}{\sqrt{t(1-t)}}$$

For the second part, $$\int_{-1}^1\frac{dt}{\sqrt{(t+1)(1-t)}}=\int_{-1}^1\frac{dt}{\sqrt{1-t^2}}=\arcsin t\Big|_{-1}^1=\pi$$

They can be combined into one step, but the rescaling substitution to [0,1] is easier to write down.

Here is an intuitive approach which may or may not be helpful, depending on how rigorous a solution you need.

First I will assume that you want a real integral, and that the question should be $$\int_a^b\frac{dx}{\sqrt{(x-a)(b-x)}}$$ as various people have suggested in comments. Note that we have two problems (though they are both of the same kind) since the integrand is unbounded as $x\to a^+$ and as $x\to b^-$.

If $x\to a^+$ then $1/\sqrt{b-x}$ is more or less a (finite, non-zero) constant. So we can tell whether the integral converges or not by considering $$\int_a^{a+\varepsilon}\frac{dx}{\sqrt{x-a}} =\lim_{c\to a^+}\int_c^{a+\varepsilon}(x-a)^{-1/2}\,dx =\lim_{c\to a^+}\bigl(2\sqrt{\varepsilon}-2\sqrt{c-a}\bigr) =2\sqrt{\varepsilon}\,.$$ Since the limit exists, the integral converges. You can deal with the problem as $x\to b^-$ in the same way.