Let I = ∫√{1 - (x - 1)² } dx

Let x - 1 = sin u

dx = cos u du

From equation 1

      I = ∫√{1 - sin² u}cos u du

=> I = cos u*cos u du

=> I = ∫sin² u du

=>  I = ∫{1 + cos 2u}/2 du

=> I = {u + (sin 2u)/2}/2 + C .............2

Again x - 1 = sin u

=> u = sin^-1 (x - 1)

From equation 2, we get

=> I = {u + (sin 2u)/2}/2 + C

=> I = [sin^-1 (x - 1) + {sin 2(sin^-1 (x - 1)}/2]/2 + C

=> I = (1/2)*sin^-1 (x - 1) + {sin 2(sin^-1 (x - 1)}/4 + C

So, ∫√{1 - (x - 1)² } dx = (1/2)*sin^-1 (x - 1) + {sin 2(sin^-1 (x - 1)}/4 + C