`int_(0)^(pi) sqrt(1+4 "sin"^(2) x/2-4 "sin"x/2) dx` का मान कीजिए।

Explanation

`int_(0)^(pi) sqrt(1+4 "sin"^(2) x/2 -4 "sin" x/2) dx`
`=int_(0)^(pi) sqrt((1-2 "sin"x/2)^(2)) dx`
`=int_(0)^(pi)|1-2 "sin" x/2| dx`
`=int_(0)^(pi//3) (1-2 "sin"x/2) dx - int_(pi//3)^(pi) (1-2"sin" x/2) dx`
`=[x+4 "cos"x/2]_(0)^(pi/3)-[x+4"cos" x/2]_(pi//3)^(pi)`
`=[pi/3-0+4 "cos" pi/6-4 "cos" 0/2]-[pi-pi/3+4 "cos" pi/2- 4 "cos" pi/6]`
`=[pi/3+2sqrt(3)-4]-[2 pi/3-2sqrt(3)]`
`=pi/3 +2 sqrt(3) -4-(2 pi)/3+2 sqrt(3)`
`=-pi/3 +4sqrt(3) -4`.