Does the integral $\int_{a}^{b}\frac{dx}{\sqrt{(x-a)(x-b)}}$ exist? What is the result of this integral $\displaystyle\int_{a}^{b}\dfrac{dx}{\sqrt{(x-a)(x-b)}}$ ? I have tried many possibilities like letting $\sqrt{(x-a)(x-b)}$=u or trying to make the denominator express as a difference of two sqares but nothing worked.
 A: 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.
A: 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.
A: Setting $a=0$ and $b=1$, then 
\begin{align}
\int_0^1 {{x^{ - {\textstyle{1 \over 2}}}}{{\left( {1 - x} \right)}^{ - {\textstyle{1 \over 2}}}}dx}  = \int_0^1 {{x^{{\textstyle{1 \over 2}} - 1}}{{\left( {1 - x} \right)}^{{\textstyle{1 \over 2}} - 1}}dx}  = B\left( {\frac{1}{2},\frac{1}{2}} \right) = {\Gamma ^2}\left( {\frac{1}{2}} \right) = \pi
\end{align}
