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
