Evaluate : `intdx/sqrt((x-1)(2-x))` by the substitution `x=1+sin^2theta`. Hence, find the value of `int_1^2dx/sqrt((x-1)(2-x))`

Options

A. `pi/2`
B. `pi/4`
C. `pi`
D. `2pi`

Answer

C

Explanation

Let `I = int_(1)^(2) (dx)/(sqrt((x-1)(2-x))) = int_(1)^(2)(dx)/(sqrt(2x-x^(2) - 2 + x))`
`= int_(1)^(2) (dx)/(sqrt(-(x^(2) - 3x + 2)))`
`= int_(1)^(2)(dx)/(sqrt(-[x^(2) - 2.(3)/(2)x+ ((3)/(2))^(2)+ 2 - (9)/(4)]))`
`= int_(1)^(2) (dx)/(sqrt(-{(x -
(3)/(2))^(2) - ((1)/(2))^(2)}))`
`= int_(1)^(2) (dx)/(sqrt(((1)/(2))^(2)- (x - (3)/(2))^(2))) = [sin^(-1)((x - (3)/(2))/((1)/(2)))]_(1)^(2)`
= `[sin^(-1)(2x - 3)]_(1)^(2) = sin^(-1)1 - sin^(-1)(-1)`
= `(pi)/(2) + (pi)/(2) " " [therefore sin'(pi)/(2) = 1 "and sin" '(-theta) = - sin theta ]`
= `pi`