Question

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

Answer 1

\[\text{Let I } = \int\frac{dx}{\sqrt{\left( x - \alpha \right) \left( \beta - x \right)}}\]

\[ = \int\frac{dx}{\sqrt{\  β x - x^2 -  αβ+ α x }}\]

\[ = \int\frac{dx}{\sqrt{- x^2 + \left( \alpha + \beta \right) x - \alpha\beta}}\]

\[ = \int\frac{dx}{\sqrt{- \left[ x^2 - \left( \alpha + \beta \right) x + \alpha\beta \right]}}\]

\[ = \int\frac{dx}{\sqrt{- \left[ x^2 - \left( \alpha + \beta \right) x + \left( \frac{\alpha + \beta}{2} \right)^2 - \left( \frac{\alpha + \beta}{2} \right)^2 + \alpha\beta \right]}}\]

\[ = \int\frac{dx}{\sqrt{- \left\{ x - \left( \frac{\alpha + \beta}{2} \right) \right\}^2 + \left( \frac{\alpha + \beta}{2} \right)^2 - \alpha\beta}}\]

\[ = \int\frac{dx}{\sqrt{- \left[ x - \left( \frac{\alpha + \beta}{2} \right) \right]^2 + \frac{\left( \alpha + \beta \right)^2 - 4\alpha\beta}{4}}}\]

\[ = \int\frac{dx}{\sqrt{- \left[ x - \left( \frac{\alpha + \beta}{2} \right) \right]^2 + \left( \frac{\alpha - \beta}{2} \right)^2}}\]

\[ = \int\frac{dx}{\sqrt{\left( \frac{\alpha - \beta}{2} \right)^2 - \left( x - \left( \frac{\alpha + \beta}{2} \right) \right)^2}}\]

\[ = \sin^{- 1} \left[ \frac{x - \left( \frac{\alpha + \beta}{2} \right)}{\frac{\alpha - \beta}{2}} \right] + C\]

\[ = \sin^{- 1} \left( \frac{2x - \alpha - \beta}{\alpha - \beta} \right) + C\]