The integral `int_(0)^(pi)sqrt(1+4"sin"^(2)x/2-4"sin"x/2)dx` is equals to (a)`pi-4` (b)`(2pi)/3-4-sqrt(3)` (c)`(2pi)/3-4-sqrt(3)` (d)`4sqrt(3)-4-(pi)/3`

Options

A. `4sqrt(3)-4`
B. `4sqrt(3)-4-(pi)/(3)`
C. `pi-4`
D. `(2pi)/(3)-4-4sqrt(3)`

Answer

B

Explanation

Let `I=overset(pi)underset(0)int sqrt(1+4"sin"^(2)(x)/(2)-4 "sin"(x)/(2))dx`. Then,
`I=overset(pi)underset(0)int sqrt((1-2"sin"(x)/(2))^(2))dx`
`rArr I=overset(pi)underset(0)int|1-2 "sin"(x)/(2)|dx" "[:'sqrt(x^(2))=|x|]`
`rArr I=overset(pi//3)underset(0)int |1-2 "sin"(x)/(2)|dx+overset(pi)underset(pi//3)int |1-2"sin"(x)/(2)|dx`
`rArr I=overset(pi//3)underset(0)int (1-2"sin"(x)/(2))dx+overset(pi)underset(pi//3)int (2"sin"(x)/(2)-1)dx`
`rArr I=[x+4"cos"(x)/(2)]_(0)^(pi//3)+[-4"cos"(x)/(2)-x]_(pi//3)^(pi)`
`rArr I=((pi)/(3)+4"cos"(pi)/(6)-4)+(0-pi+4"cos"(pi)/(6)+(pi)/(3))`
`rArr I=-(pi)/(3)+4sqrt(3)-4`