Answer is:
#6sin^-1((x - 1)/2) + c#
To me the me the easiest way to evaluate this is to use a substitution that takes #sinx#
Because #int1/(sqrt(1 - x^2))dx = sin^-1(x) + c#
Right?
To Evaluate : # 6int1/(sqrt(4 - (x - 1)^2))dx#
step1:
let #x - 1 = 2sintheta => (dx)/(d theta) = 2cos theta => dx = 2costheta*d theta#
so that,
#(x - 1)^2 = 4sin^2theta#
#4 - (x - 1)^2 = 4 - 4sin^2theta#
#sqrt(4 - (x - 1)^2) = sqrt(4 - 4sin^2(theta)) = sqrt(4(1 - sin^2theta)) = sqrt(4cos^2theta)#
# = 2costheta#
#1/(sqrt(4 - (x - 1)^2)) = 1/(2costheta)#
#=> 6int1/(sqrt(4 - (x - 1)^2))dx = 6int1/(2costheta)*2costheta*d theta#
#= 6intd theta = 6*theta + c#
Remember that #x - 1 = 2sintheta => theta = sin^-1((x - 1)/2)#
#=> 6theta + c = 6*sin^-1((x - 1)/2) + c#
