Evaluating an improper integral with radical in denominator For $a,b\in\mathbb{R}$ and $a<b$, I'd like to evaluate the following improper integral
$$
\int_a^b \frac{dt}{\sqrt{(b-t)(t-a)}}
$$
Since I'd like to evaluate the integral rather than just make sure it converges, the comparison test isn't particularly useful. I started by expanding the terms under the radical, and then I attempted to complete the square. This left me with the following
$$
\int_a^b \frac{dt}{\sqrt{\left(t-\frac{a+b}{2}\right)^2 - \left(\frac{a+b}{2}\right)^2+ab}}
$$
In order for completing the square to be useful, I'd need to be able to make $ab-\left(\frac{a+b}{2}\right)^2$ into a square. I'll just sweep everything under the rug and say
$$
ab-\left(\frac{a+b}{2}\right)^2 = c^2 
$$
for some $c\in\mathbb{R}$. Similarly, I'll let
$$
u^2=\left(t-\frac{a+b}{2}\right)^2
$$
Then my resulting integral has the form
$$
\int_a^b \frac{dt}{\sqrt{u^2-c^2}}
$$
I think at this point, I'm supposed to use a trig substitution to finish evaluating the integral. However, because of the substitutions I made above, I'm not really sure how to go about doing the actual substitution. Is there a better way to go about evaluating this integral, or do I just need to continue grinding it out?
 A: Alternative approach:
$$\frac{1}{\sqrt{(a-x)(b-x)}}=\left(\frac{1}{\sqrt{a-x}}+\frac{1}{\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)=-2\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)$$
Thus:
$$\int_a^b \frac{1}{\sqrt{(a-x)(b-x)}}\;dx=-2\int_a^b\left(\frac{-1}{2\sqrt{a-x}}+\frac{-1}{2\sqrt{b-x}}\right)\left(\frac{1}{\sqrt{a-x}+\sqrt{b-x}}\right)\;dx$$
$$=-2\ln(\sqrt{a-x}+\sqrt{b-x})|_a^b=\ln\left(\frac{b-a}{a-b}\right)=i\pi$$
Edit: noticed there has been a sign change under the square root in OP. This is equivalent to multiplying the above by $1/i$, so the result is now $\pi$, as Michael Hardy has shown.
A: \begin{align}
u & = \frac{t-a}{b-a} \\[8pt]
dt & = (b-a)du \\[8pt]
\sqrt{(b-t)(t-a)} & = (b-a)\sqrt{u(1-u)}
\end{align}
As $t$ goes from $a$ to $b$, then $u$ goes from $0$ to $1$.  So
$$
\int_a^b\frac{dt}{\sqrt{(b-t)(t-a}} = \int_0^1 \frac{du}{\sqrt{u(1-u)}} = \int_0^1 u^{1/2-1} (1-u)^{1/2-1}\,du = B\left(\frac12,\frac12\right).
$$
$$
=\frac{\Gamma\left(\frac12\right)\Gamma\left(\frac12\right)}{\Gamma\left(\frac12+\frac12\right)} = \pi.
$$
Later addendum: A standard identity says that if
$$
B(\alpha,\beta)=\int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\,dx
$$
and
$$
\Gamma(\alpha)=\int_0^\infty x^{\alpha-1} e^{-x}\,dx
$$
then
$$
B(\alpha,\beta)=\frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}.
$$
A proof is as follows.
$$
\Gamma(\alpha)\Gamma(\beta) = \int_0^\infty u^{\alpha-1}e^{-u}\,du \int_0^\infty v^{\beta-1}e^{-v}\,dv =\int_0^\infty\int_0^\infty u^{\alpha-1}v^{\beta-1}e^{-(u+v)}\,du\,dv.\tag{1}
$$
Now let
$$x=\frac{u}{u+v}, \qquad y=u+v,$$
$$dx\,dy=\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv = \frac{du\,dv}{u+v} = \frac{du\,dv}{y}$$
$$
u = xy,\qquad v = (1-x)y.
$$
So $(1)$ becomes
$$
\int_0^\infty\int_0^1 (xy)^{\alpha-1} ((1-x)y)^{\beta-1} e^{-y} y\,dx\,dy
$$
$$
= \int_0^1 x^{\alpha-1}(1-x)^{\beta-1}\,dx\cdot\int_0^\infty y^{\alpha+\beta-1} e^{-y}\,dy = B(\alpha,\beta)\Gamma(\alpha+\beta).
$$
