Question

$\int \frac{1}{\sqrt{\left(x-\mathrm{\alpha }\right)\left(\mathrm{\beta }-x\right)}}dx,\left(\mathrm{\beta }>\mathrm{\alpha }\right)$

Answer 1

$$\begin{gathered} \mathrm{Let}I=\int \frac{dx}{\sqrt{\left(x-\alpha \right)\left(\beta -x\right)}} \\ =\int \frac{dx}{\sqrt{\beta x-x^2-\alpha \beta +\alpha x}} \\ =\int \frac{dx}{\sqrt{-x^2+\left(\alpha +\beta \right)x-\alpha \beta }} \\ =\int \frac{dx}{\sqrt{-\left[x^2-\left(\alpha +\beta \right)x+\alpha \beta \right]}} \\ =\int \frac{dx}{\sqrt{-\left[x^2-\left(\alpha +\beta \right)x+{\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}^2-{\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}^2+\alpha \beta \right]}} \\ =\int \frac{dx}{\sqrt{-{\left\{x-\left({\displaystyle \frac{\alpha +\beta }{2}}\right)\right\}}^2+{\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}^2-\alpha \beta }} \\ =\int \frac{dx}{\sqrt{-{\left[x-\left({\displaystyle \frac{\alpha +\beta }{2}}\right)\right]}^2+{\displaystyle \frac{{\left(\alpha +\beta \right)}^2-4\alpha \beta }{4}}}} \\ =\int \frac{dx}{\sqrt{-{\left[x-\left({\displaystyle \frac{\alpha +\beta }{2}}\right)\right]}^2+{\left({\displaystyle \frac{\alpha -\beta }{2}}\right)}^2}} \\ =\int \frac{dx}{\sqrt{{\left({\displaystyle \frac{\alpha -\beta }{2}}\right)}^2-{\left(x-\left({\displaystyle \frac{\alpha +\beta }{2}}\right)\right)}^2}} \\ ={\sin }^{-1}\left[\frac{x-\left({\displaystyle \frac{\alpha +\beta }{2}}\right)}{{\displaystyle \frac{\alpha -\beta }{2}}}\right]+C \\ ={\sin }^{-1}\left(\frac{2x-\alpha -\beta }{\alpha -\beta }\right)+C \end{gathered}$$