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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.

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    $\begingroup$ First: Note that as stated the integral (if it exists) must be imaginary owing to the denominator. To get rid of that, change the square root to $\sqrt{(x-a)(b-x)}$. Second: Try a simpler case first---say, $a=-b=1$. Do you see a useful transformation? $\endgroup$ Commented Aug 4, 2014 at 5:16
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    $\begingroup$ Assuming $a\leq b$, the argument of the square root is negative for any $x\in(a,b)$, hence the integrand is not a real function. Regarding the integrand function as a complex-valued function, you have to choice a determination for the square root. $\endgroup$ Commented Aug 4, 2014 at 5:20
  • $\begingroup$ Try the substitution $x=a\sin^2y+b\cos^2y$. I think it will work. $\endgroup$
    – Bumblebee
    Commented Aug 4, 2014 at 9:25

3 Answers 3

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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.

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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.

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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}

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