let-give-gt-0-find-the-value-of-0-1-dx-1-x-1-x-

Question Number 33120 by abdo imad last updated on 10/Apr/18

l e t g i v e α > 0 f i n d t h e v a l u e o f \int_{0}^{1} \frac{d x}{\sqrt{(1 - x) (1 + α x)}} .

Answered by MJS last updated on 11/Apr/18

(1 - x) (1 + α x) = - α x^{2} + (α - 1) x + 1

\int \frac{d x}{\sqrt{{a x}^{2} + b x + c}} = - \frac{1}{\sqrt{- a}} \sin^{- 1} \frac{2 a x + b}{\sqrt{b^{2} - 4 a c}}; a < 0

\underset{0}{\int^{1}} \frac{d x}{\sqrt{- α x^{2} + (α - 1) x + 1}} = {[- \frac{1}{\sqrt{α}} \sin^{- 1} \frac{- 2 α x + α - 1}{\sqrt{{(α - 1)}^{2} + 4 α}}]}_{0}^{1} =

= - \frac{1}{\sqrt{α}} {[\sin^{- 1} \frac{- 2 α x + α - 1}{α + 1}]}_{0}^{1} =

= - \frac{1}{\sqrt{α}} (\sin^{- 1} \frac{- α - 1}{α + 1} - \sin^{- 1} \frac{α - 1}{α + 1}) =

= \frac{1}{\sqrt{α}} (\sin^{- 1} 1 + \sin^{- 1} \frac{α - 1}{α + 1}) =

= \frac{1}{\sqrt{α}} (\frac{π}{2} + \sin^{- 1} \frac{α - 1}{α + 1})

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