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

let give α>0 find the value of ∫_0 ^1 (dx/(√((1−x)(1+αx)))) .

letgiveα>0findthevalueof∫01dx(1−x)(1+αx).

Answered by MJS last updated on 11/Apr/18

(1−x)(1+αx)=−αx^2 +(α−1)x+1 ∫(dx/(√(ax^2 +bx+c)))=−(1/(√(−a)))sin^(−1) ((2ax+b)/(√(b^2 −4ac))); a<0 ∫_0 ^1 (dx/(√(−αx^2 +(α−1)x+1)))=[−(1/(√α))sin^(−1) ((−2αx+α−1)/(√((α−1)^2 +4α)))]_0 ^1 = =−(1/(√α))[sin^(−1) ((−2αx+α−1)/(α+1))]_0 ^1 = =−(1/(√α))(sin^(−1) ((−α−1)/(α+1))−sin^(−1) ((α−1)/(α+1)))= =(1/(√α))(sin^(−1) 1+sin^(−1) ((α−1)/(α+1)))= =(1/(√α))((π/2)+sin^(−1) ((α−1)/(α+1)))

(1−x)(1+αx)=−αx2+(α−1)x+1∫dxax2+bx+c=−1−asin−12ax+bb2−4ac;a<0∫10dx−αx2+(α−1)x+1=[−1αsin−1−2αx+α−1(α−1)2+4α]01==−1α[sin−1−2αx+α−1α+1]01==−1α(sin−1−α−1α+1−sin−1α−1α+1)==1α(sin−11+sin−1α−1α+1)==1α(π2+sin−1α−1α+1)

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