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find-0-1-dx-x-1-x-




Question Number 42435 by maxmathsup by imad last updated on 25/Aug/18
find  ∫_0 ^1    (dx/( (√x) +(√(1−x))))
find01dxx+1x
Commented by maxmathsup by imad last updated on 25/Aug/18
let I = ∫_0 ^1     (dx/( (√x) +(√(1−x))))   changement x=sin^2 t give  I =  ∫_0 ^(π/2)     ((2 sint cost dt)/(sint +cost)) = ∫_0 ^(π/2)    ((sin^2 t +cos^2 t +2sint cost −1)/(sint +cost))dt  = ∫_0 ^(π/2)    (((sint +cost)^2 −1)/(sint +cost)) dt =∫_0 ^(π/2) (sint +cost)dt −∫_0 ^(π/2)   (dt/(sint +cost))  =[sint −cost]_0 ^(π/2)   − ∫_0 ^(π/2)    (dt/(sint +cost))  =2 −∫_0 ^(π/2)    (dt/(sint +cost))  changement tan((t/2))=u give  ∫_0 ^(π/2)    (dt/(sint +cost)) =∫_0 ^1      (1/(((2u)/(1+u^2 )) +((1−u^2 )/(1+u^2 ))))  ((2du)/(1+u^2 )) = 2 ∫_0 ^1     (du/(2u+1−u^2 ))  =−2  ∫_0 ^1    (du/(u^2 −2u −1))  =−2  ∫_0 ^1     (du/((u−1)^2 −1)) =_(u−1 =−α)  −2  ∫_1 ^0     ((−dα)/(α^2 −1))      =−2 ∫_0 ^1    (dα/(α^2 −1)) =− ∫_0 ^1    ((1/(α−1))−(1/(α+1))) =−[ln∣((α−1)/(α +1))∣_0 ^1  = +∞ so this  integral diverges !..
letI=01dxx+1xchangementx=sin2tgiveI=0π22sintcostdtsint+cost=0π2sin2t+cos2t+2sintcost1sint+costdt=0π2(sint+cost)21sint+costdt=0π2(sint+cost)dt0π2dtsint+cost=[sintcost]0π20π2dtsint+cost=20π2dtsint+costchangementtan(t2)=ugive0π2dtsint+cost=0112u1+u2+1u21+u22du1+u2=201du2u+1u2=201duu22u1=201du(u1)21=u1=α210dαα21=201dαα21=01(1α11α+1)=[lnα1α+101=+sothisintegraldiverges!..
Answered by tanmay.chaudhury50@gmail.com last updated on 25/Aug/18
I=  x=sin^2 α   dx=2sinαcosα  ∫_0 ^(Π/2) ((2sinαcosα)/(sinα+cosα))dα  ∫_0 ^(Π/2) ((sin^2 α+cos^2 α+2sinαcosα−1)/(sinα+cosα))dα  ∫_0 ^(Π/2) sinα+cosα dα−∫_0 ^(Π/2) (dα/(sinα+cosα))  ∫_0 ^(Π/2) sinα+cosα dα−∫_0 ^(Π/2) ((sec^2 (α/2))/(2tan(α/2)+1−tan^2 (α/2)))dα  ∫_0 ^(Π/2) sinα+cosα dα+2∫_0 ^(Π/2) ((d(tan(α/2)))/(tan^2 (α/2)−2tan(α/2)+1−2))  ∫_0 ^(Π/2) sinα+cosα dα+2∫_0 ^(Π/2) ((d(tan(α/2)))/((tan(α/2)−1)^2 −1))  ∫_0 ^(Π/2) sinα+cosα dα−2∫_0 ^(Π/2) ((d(tan(α/2)))/(1−(tan(α/2)−1)^2 ))  ∫_0 ^(Π/2)  sinα+cosα dα−2∫_0 ^(Π/2) ((d(tan(α/2) −1))/((tan(α/2)−1)^2 −1))→∫(dx/(x^2 −a^2 ))  =∣−cosα+sinα∣_0 ^(Π/2) −2×(1/2)ln∣((tan(α/2)−1−1)/(tan(α/2)−1+1))∣_0 ^(Π/2)   ={(−0+1)−(−1+0)}−{ln∣((1−1−1)/(1−1+1))∣−ln∣((−2)/0)∣}  =2−{ln∣−1∣−ln∣−∞∣}
I=x=sin2αdx=2sinαcosα0Π22sinαcosαsinα+cosαdα0Π2sin2α+cos2α+2sinαcosα1sinα+cosαdα0Π2sinα+cosαdα0Π2dαsinα+cosα0Π2sinα+cosαdα0Π2sec2α22tanα2+1tan2α2dα0Π2sinα+cosαdα+20Π2d(tanα2)tan2α22tanα2+120Π2sinα+cosαdα+20Π2d(tanα2)(tanα21)210Π2sinα+cosαdα20Π2d(tanα2)1(tanα21)20Π2sinα+cosαdα20Π2d(tanα21)(tanα21)21dxx2a2=∣cosα+sinα0Π22×12lntanα211tanα21+10Π2={(0+1)(1+0)}{ln11111+1ln20}=2{ln1ln}

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