Funny math problem to calculate the Pin Number for a credit card.

[Albert Chan]

(08-11-2021 03:40 PM)Albert Chan Wrote:  I2 = ∫(1/√(x^2-3x+2), x=0..1)            // let y = 2*(x-3/2)=2x-3, dy = 2 dx
    = ∫(1/√(y^2-1), y=-3 .. -1)             // ∫(1/√(y^2-1), y) = -ln(abs(-y+sqrt(y^2-1)))
    = ln(2*√(2)+3)

Another way is to map y=2*x-3 for the full integral, then solve by integration by parts
Note that above quoted expression, map from x to y, numerator of 1 stayed 1.

XCAS> e2r(r2e([3,-1,2,-4], (y+3)/2), y)       → [3/8, 25/8, 77/8, 55/8]

I = ∫((3y^3 + 25y^2 + 77y + 55) / √(y^2-1), y=-3 .. -1) / 8

Y0 = ∫(1/√(y^2-1),y) = -ln(abs(-y+√(y^2-1)))
Y1 = ∫(y/√(y^2-1),y) = ∫d(√(y^2-1)) = √(y^2-1)

Y2 = ∫(y^2/√(y^2-1),y) = ∫(y*d(Y1)) = y*Y1 - (Y2-Y0)
2*Y2 = y*Y1 + Y0

Y3 = ∫(y^3/√(y^2-1),y) = ∫(y^2*d(Y1)) = y^2*Y1 - 2*(Y3-Y1)
3*Y3 = (y^2+2)*Y1

3*Y3 + 25*Y2 + 77*Y1 + 55*Y0
= (y^2+2)*Y1 + 25/2*(y*Y1+Y0) + 77*Y1 + 55*Y0
= (y^2+25/2*y+79)*Y1 + 135/2*Y0

I = preval((y^2+25/2*y+79)*√(y^2-1) + 135/2*(-ln(abs(-y+√(y^2-1)))), -3, -1, y) / 8
  = -101/8*√(2) + 135/16*ln(2*√(2)+3)
  ≈ -2.98126694401